A+B: Difference between revisions
jq |
Using a let block for such a trivial program was unneeded. This is similar to the Arc version. |
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=={{header|Scheme}}== |
=={{header|Scheme}}== |
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<lang scheme> |
<lang scheme>(display (+ (read) (read)))</lang> |
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(let* ((x (read)) |
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(y (read))) |
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(write (+ x y))) |
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</lang> |
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=={{header|sed}}== |
=={{header|sed}}== |
Revision as of 22:46, 24 August 2014
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
A+B - in programming contests, classic problem, which is given so contestants can gain familiarity with the online judging system being used.
Problem statement
Given 2 integer numbers, A and B. One needs to find their sum.
- Input data
- Two integer numbers are written in the input stream, separated by space.
- Output data
- The required output is one integer: the sum of A and B.
- Example:
Input Output 2 2 4 3 2 5
0815
<lang 0815>|x|+%</lang>
ABAP
<lang ABAP>report z_sum_a_b. data: lv_output type i. selection-screen begin of block input.
parameters: p_first type i, p_second type i.
selection-screen end of block input.
at selection-screen output.
%_p_first_%_app_%-text = 'First Number: '. %_p_second_%_app_%-text = 'Second Number: '.
start-of-selection.
lv_output = p_first + p_second. write : / lv_output.</lang>
Ada
<lang Ada>-- Standard I/O Streams
with Ada.Integer_Text_Io; procedure APlusB is
A, B : Integer;
begin
Ada.Integer_Text_Io.Get (Item => A); Ada.Integer_Text_Io.Get (Item => B); Ada.Integer_Text_Io.Put (A+B);
end APlusB;</lang> Using appropriate user defined types: <lang Ada>with Ada.Text_IO;
procedure A_Plus_B is
type Small_Integers is range -2_000 .. +2_000; subtype Input_Values is Small_Integers range -1_000 .. +1_000; package IO is new Ada.Text_IO.Integer_IO (Num => Small_Integers); A, B : Input_Values;
begin
IO.Get (A); IO.Get (B); IO.Put (A + B, Width => 4, Base => 10);
end A_Plus_B;</lang>
Aime
<lang aime>file f; list l;
f_affix(f, "/dev/stdin"); f_list(f, l, 0); o_integer(atoi(l_q_text(l, 0)) + atoi(l_q_text(l, 1))); o_newline();</lang>
ALGOL 68
Console
<lang algol68>print((read int + read int))</lang> Input:
1 2
Output:
+3
File
<lang algol68>open(stand in, "input.txt", stand in channel); open(stand out, "output.txt", stand out channel); print((read int + read int))</lang> Input "input.txt":
3 4
Output "output.txt":
+7
ANTLR
![aplusb](http://static.miraheze.org/rosettacodewiki/5/59/Aplusb.jpg)
![aplusb](http://static.miraheze.org/rosettacodewiki/5/52/Num.png)
Java
<lang> grammar aplusb ;
options { language = Java; }
aplusb : (WS* e1=Num WS+ e2=Num NEWLINE {System.out.println($e1.text + " + " + $e2.text + " = " + (Integer.parseInt($e1.text) + Integer.parseInt($e2.text)));})+ ; Num : '-'?('0'..'9')+ ; WS : (' ' | '\t') ; NEWLINE : WS* '\r'? '\n'
;
</lang> Produces:
>java Test 1 2 23 89 13 567 -75 6 -75 -29 ^Z 1 + 2 = 3 23 + 89 = 112 13 + 567 = 580 -75 + 6 = -69 -75 + -29 = -104
Arc
<lang Arc> (prn (+ (read)
(read)))
</lang>
Argile
<lang Argile>(: Standard input-output streams :) use std, array Cfunc scanf "%d%d" (&val int a) (&val int b) printf "%d\n" (a + b)</lang> <lang Argile>(: Input file : input.txt :) (: Output file: output.txt :) use std, array let in = fopen "input.txt" "r" let out = fopen "output.txt" "w" let int x, y. Cfunc fscanf in "%d%d" (&x) (&y) (:fscanf not yet defined in std.arg:) fprintf out "%d\n" (x+y) fclose in fclose out</lang>
ARM Assembly
Todo: -need to print numbers w/o the leading 0's. Replace them with spaces, so alignment is still the same.
Read two strings from stdin, convert to integers calculate their sum, print to stdout. A valid int is a value between -2^31 (-2147483647) and 2^31-1 (2147483647). We do not allow -2147483648 as input, but it is a valid result. E.g. -1 -2147483647. Maximum number of digits is 10. Leading 0's are counted as number length. We read signed values. We ignore leading '+'s and allow '-' for negative values. If multiple plus or minus signs precede a number, only the last one counts. minval and maxval can be used to specify any valid range, (e.g. -1000 and +1000). The range is inclusive. If 0 is specified for both ranges, range checks are not done.
Tested on RaspberryPi model B (GNU/Linux, ARMv6) Save in ab.S Build with: <lang ARM_Assembly>as -o ab.o ab.S ld -o a.out ab.o</lang>
<lang ARM_Assembly>.data
.align 2 .code 32
.section .rodata
.align 2 .code 32
overflow_msg: .ascii "Invalid number. Overflow.\n" overflow_msglen = . - overflow_msg bad_input_msg: .ascii "Invalid input. NaN.\n" bad_input_msglen = . - bad_input_msg range_err_msg: .ascii "Value out of range.\n" range_err_msglen = . - range_err_msg io_error_msg: .ascii "I/O error.\n" io_error_msglen = . - range_err_msg
sys_exit = 1 sys_read = 3 sys_write = 4 max_rd_buf = 14 lf = 10 m10_9 = 0x3b9aca00 maxval = 1000 minval = -1000
.text
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ void main() @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type _start STT_FUNC .global _start
_start:
stmfd sp!, {r4,r5,lr}
.read_lhs:
ldr r0, =max_rd_buf bl readint mov r4, r0 bl printint mov r0, r4 bl range_check
.read_rhs:
ldr r0, =max_rd_buf bl readint mov r5, r0 bl printint mov r0, r5 bl range_check
.sum_and_print:
adds r0, r4, r5 bvs overflow bl printint
.main_exit:
mov r0, #0 bl exit ldmfd sp!, {r4,r5,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ Read from stdin until we encounter a non-digit, or we have read bytes2rd digits. @@ Ignore leading spaces. @@ Return value to the caller converted to a signed int. @@ We read positive values, but if we read a leading '-' sign, we convert the @@ return value to two's complement. @@ The argument is max number of bytes to read from stdin. @@ int readint(int bytes2rd) @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type readint STT_FUNC .global readint
readint:
stmfd sp!, {r4,r5,r6,r7,lr} @@@@@@@@@@@@@@@ @@ r0 : #0 for stdin arg to read. @@ r1 : ptr to current pos in local buffer. @@ r2 : #1 to read one byte at a time. @@ r3,r7 : tmp. @@ r4 : number of bytes read. @@ r5 : value of current byte. @@ r6 : 0 while we are reading leading spaces. @@@@@@@@@@@@@@@ sub sp, sp, r0 mov r1, sp mov r3, #0 push {r3} @ sp,#4: local var @isnegative. return in r1. Default value is 0/false. Positive number. push {r0} @ sp,#0: local var @maxbytes. const. mov r2, #1 mov r4, #0
mov r6, #0 b .rd
@ we get here if r6 is 0. @ if space, goto .rd. @ else set r6 to 1 and goto .noleading. .leadchk:
mov r0, r5 bl isspace cmp r0, #1 beq .rd
.sign_chk:
mov r0, r5 push {r1} bl issign cmp r0, #1 streq r0, [sp,#8] @ sp,#4 + 4 for the pushed r1. movhi r1, #0 strhi r1, [sp,#8] @ sp,#4 + 4 for the pushed r1. pop {r1} bhs .rd
mov r6, #1 b .noleading
.rd:
mov r0, #0 bl read cmp r0, #1 bne .sum_digits_eof @ eof mov r5, #0 ldrb r5, [r1] cmp r6, #0 beq .leadchk
.noleading:
mov r0, r5 bl isdigit cmp r0, #1 bne .sum_digits_nan @ r5 is non-digit
add r4, r4, #1 add r1, r1, #1 @ max chars to read is received in arg[0], stored in local var at sp. @ Only 10 can be valid, so the default of 12 leaves space for separator. ldr r3, [sp] cmp r4, r3 beq .sum_digits_maxrd @ max bytes read. b .rd
@@@@@@@@@@@@@@@ @ We have read r4 (0..arg[0](default 12)) digits when we get here. Go through them @ and add/mul them together to calculate a number. @ We multiply and add the digits in reverse order to simplify the multiplication. @@@@@@@@@@@@@@@ @ r0: return value. @ r1: local variable for read buffer. @ r2: tmp for conversion. @ r3,r6,r7: tmp @ r4: number of chars we have read. @ r5: multiplier 1,10,100. @@@@@@@@@@@@@@@
.sum_digits_nan:
mov r0, r5 bl isspace cmp r0, #1 bne bad_input
.sum_digits_maxrd: .sum_digits_eof:
mov r0, #0 mov r5, #1
.count:
cmp r4, #0 beq .readint_ret sub r4, r4, #1 sub r1, #1 ldrb r2, [r1] sub r2, r2, #48 mov r3, r2
@ multiply r3 (char value of digit) with r5 (multiplier). @ possible overflow. @ MI means negative. @ smulls multiples two signed 32 bit vals and returns a 64 bit result. @ If we get anything in r7, the value has overflowed. @ having r2[31] set is overflow too. smulls r2, r7, r3, r5 cmp r7, #0 bne overflow cmp r2, #0 bmi overflow
@@ possible overflow. adds r0, r0, r2 bvs overflow bmi overflow
@@ end of array check. @@ check is needed here too, for large numbers, since 10 billion is not a valid 32 bit val. cmp r4, #0 beq .readint_ret
@@ multiple multiplier by 10. @@ possible overflow. @@ too many digits is input. happens if input is more than 10 digits. mov r3, #10 mov r6, r5 smulls r5, r7, r3, r6 cmp r7, #0 bne overflow cmp r5, #0 bmi overflow b .count
.readint_ret:
ldr r1, [sp,#4] @ read isnegative value. cmp r1, #0 rsbne r0, r0, #0 pop {r2} add sp, sp, #4 add sp, sp, r2 ldmfd sp!, {r4,r5,r6,r7,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ int isdigit(int) @@ #48..#57 ascii range for '0'..'9'. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type isdigit STT_FUNC .global isdigit
isdigit:
stmfd sp!, {r1,lr} cmp r0, #48 blo .o_range cmp r0, #57 bhi .o_range mov r0, #1 ldmfd sp!, {r1,pc}
.o_range:
mov r0, #0 ldmfd sp!, {r1,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ int isspace(int) @@ ascii space = 32, tab = 9, newline 10, cr = 13. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type isspace STT_FUNC .global isspace
isspace:
stmfd sp!, {lr} cmp r0, #32 cmpne r0, #9 cmpne r0, #10 cmpne r0, #13 beq .is_space mov r0, #0 ldmfd sp!, {pc}
.is_space:
mov r0, #1 ldmfd sp!, {pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ Return value is 1 for '-' 2 for '+'. @@ int isspace(int) @@ '+' = 43 and '-' = 45. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type issign STT_FUNC .global issign
issign:
stmfd sp!, {lr} cmp r0, #43 beq .plus_sign cmp r0, #45 beq .minus_sign mov r0, #0 ldmfd sp!, {pc}
.plus_sign:
mov r0, #2 ldmfd sp!, {pc}
.minus_sign:
mov r0, #1 ldmfd sp!, {pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ ARGS: @@ r0 : in out arg (current int value) @@ r1 : in out arg (ptr to current pos in buffer) @@ r2 : in arg (const increment. 1000_000_000, 100_000_000, 10_000_000, 1000_000, 100_000, 10_000, 1000, 100, 10, 1.) @@ @@ r4 : tmp local. Outer scope must init to #10 and count down to #0. @@ Special case is INTMAX. Must init to 5 if r4 >= 1000_000_000 (0x3b9aca00 = m10_9). @@ r5: tmp @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type get_digit STT_FUNC .global get_digit
get_digit:
stmfd sp!, {r2,r4,r5,lr} ldr r5, =m10_9 cmp r2, r5 movlo r4, #10 movhs r4, #5
.get_digit_loop:
sub r4, #1 mul r5, r4, r2 cmp r0, r5 blo .get_digit_loop sub r0, r5 add r4, r4, #48 strb r4, [r1], #1 ldmfd sp!, {r2,r4,r5,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ A quick way to divide (numbers evenly divisible by 10) by 10. @@ Most ARM cpus don't have a divide instruction, @@ so this will always work. @@ A generic div function is long and not needed here. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
.div_r2_10:
stmfd sp!, {r0,r1,r3,lr} mov r0, #1 mov r1, #10
.find_x:
mul r3, r0, r1; cmp r3, r2 movlo r0, r3 blo .find_x mov r2, r0 ldmfd sp!, {r0,r1,r3,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
.print_neg_sign:
stmfd sp!, {r0,r1,r2,lr} @ 45 = '-' mov r1, #45 push {r1} mov r2, #1 @ r1 is ptr to our local variable (holding '-'). mov r1, sp mov r0, #1 bl write cmp r0, #0 blne io_error pop {r1} ldmfd sp!, {r0,r1,r2,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ void printint(int val) @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type printint STT_FUNC .global printint
printint:
stmfd sp!, {r4,r5,r6,lr} mov r1, #1 ands r1, r1, r0, LSR #31 rsbne r0, r0, #0 blne .print_neg_sign sub sp, sp, #20 mov r1, sp mov r3, sp
ldr r2, =m10_9
.getc_loop:
bl get_digit cmp r2, #1 beq .exit_getc_loop bl .div_r2_10 b .getc_loop
.exit_getc_loop:
ldr r0, =lf strb r0, [r1], #1
sub r2, r1, r3 mov r1, r3 mov r0, #1 bl write cmp r0, #0 blne io_error add sp, sp, #20 ldmfd sp!, {r4,r5,r6,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @@ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
range_check:
stmfd sp!, {r4,r5,lr} ldr r4, =minval ldr r5, =maxval cmp r4, #0 cmpeq r5, #0 beq .skip_range_check cmp r0, r4 bllt range_err cmp r0, r5 blgt range_err
.skip_range_check:
ldmfd sp!, {r4,r5,pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ void range_err() @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
range_err:
stmfd sp!, {lr} ldr r2, =range_err_msglen ldr r1, =range_err_msg mov r0, #2 bl write mov r0, #-1 bl exit ldmfd sp!, {pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ void overflow() @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
overflow:
stmfd sp!, {lr} ldr r2, =overflow_msglen ldr r1, =overflow_msg mov r0, #2 bl write mov r0, #-1 bl exit ldmfd sp!, { pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ void bad_input() @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
bad_input:
stmfd sp!, {lr} ldr r2, =bad_input_msglen ldr r1, =bad_input_msg mov r0, #2 bl write mov r0, #-1 bl exit ldmfd sp!, {pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ void io_error() @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2
io_error:
stmfd sp!, {lr} ldr r2, =io_error_msglen ldr r1, =io_error_msg mov r0, #2 bl write mov r0, #-1 bl exit ldmfd sp!, {pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ void exit(int) @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type _start STT_FUNC .global exit
exit:
stmfd sp!, {r7, lr} ldr r7, =sys_exit svc #0 ldmfd sp!, {r7, pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ int write(int fd,char*buf,int len) @ Return 0 if we successfully write all bytes. Otherwise return the error code. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type _start STT_FUNC .global write
write:
stmfd sp!, {r4,r7, lr} mov r4, r2
.wr_loop:
ldr r7, =sys_write svc #0 @ If r0 is negative, it is more than r4 with LO (unsigned <). cmp r0, r4 sublo r4, r0 blo .wr_loop moveq r0, #0 ldmfd sp!, {r4,r7, pc}
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ int read(int fd,char*buf,int len) @ Return number of bytes successfully read. Ignore errors. @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
.align 2 .code 32 .type _start STT_FUNC .global read
read:
stmfd sp!, {r7, lr} ldr r7, =sys_read svc #0 cmp r0, #0 movlt r0, #0 ldmfd sp!, {r7, pc}
</lang>
AutoHotkey
<lang AutoHotkey>InputBox, input , A+B, Two integer numbers`, separated by space. StringSplit, output, input, %A_Space% msgbox, % output1 . "+" . output2 "=" output1+output2</lang>
AutoIt
<lang AutoIt>;AutoIt Version: 3.2.10.0 $num = "45 54" consolewrite ("Sum of " & $num & " is: " & sum($num)) Func sum($numbers)
$numm = StringSplit($numbers," ") Return $numm[1]+$numm[$numm[0]]
EndFunc</lang>
AWK
<lang awk>{print $1 + $2}</lang>
Batch File
Prompts version <lang dos>::aplusb.cmd @echo off setlocal set /p a="A: " set /p b="B: " set /a c=a+b echo %c% endlocal</lang> All on the commandline version <lang dos>::aplusb.cmd @echo off setlocal set a=%1 set b=%2 set /a c=a+b echo %c% endlocal</lang> Formula on the command line version <lang dos>::aplusb.cmd @echo off setlocal set /a c=%~1 echo %c% endlocal</lang> Example of 'Formula on the command line version'
>aplusb 123+456 579 >aplusb "1+999" 1000
Parse the input stream version (thanks to Tom Lavedas on alt.msdos.batch.nt) <lang dos>::aplusb.cmd @echo off setlocal set /p a="Input stream: " call :add %a% echo %res% endlocal goto :eof
- add
set /a res=res+%1 shift if "%1" neq "" goto :add</lang> Example of 'parse the input stream version'
>aplusb Input stream: 1234 5678 6912 >aplusb Input stream: 123 234 345 456 567 678 789 890 4082
BASIC
<lang qbasic>DEFINT A-Z
tryagain: backhere = CSRLIN INPUT "", i$ i$ = LTRIM$(RTRIM$(i$)) where = INSTR(i$, " ") IF where THEN
a = VAL(LEFT$(i$, where - 1)) b = VAL(MID$(i$, where + 1)) c = a + b LOCATE backhere, LEN(i$) + 1 PRINT c
ELSE
GOTO tryagain
END IF</lang>
Applesoft BASIC
<lang ApplesoftBasic>10 BH = PEEK(37) 20 INPUT ""; A$ : I$ = A$ : VTAB BH : A = PEEK(40) + PEEK(41) * 256 : FOR S = 0 TO 39 : IF PEEK(A + S) = 160 THEN NEXT S : S = 0 40 IF LEN(I$) THEN IF MID$(I$, LEN(I$), 1) = " " THEN I$ = MID$(I$, 1, LEN(I$) - 1) : GOTO 40RTRIM 50 IF LEN(I$) < 3 THEN 10"TRY AGAIN 60 FOR WHERE = 1 TO LEN(I$) : IF MID$(I$, WHERE, 1) <> " " THEN NEXT WHERE : GOTO 10"TRY AGAIN 70 A% = VAL(LEFT$(I$, WHERE - 1)) 80 B% = VAL(MID$(I$, WHERE + 1, LEN(I$))) 90 C% = A% + B% 100 VTAB BH 110 HTAB LEN(A$) + 2 + S 120 PRINT C%</lang>
BASIC256
<lang basic256>dim a(2) input "Enter two numbers seperated by a space?", t$ a = explode(t$," ") print t$ + " " + (a[0] + a[1])</lang>
BBC BASIC
<lang bbc> REPEAT
hereY% = VPOS INPUT LINE "" q$ hereX% = LEN(q$) + 1 WHILE LEFT$(q$, 1) = " " q$ = MID$(q$, 2) ENDWHILE space% = INSTR(q$, " ") IF space% THEN a = VAL(LEFT$(q$, space% - 1)) b = VAL(MID$(q$, space% + 1)) PRINT TAB(hereX%, hereY%) ; a + b ENDIF UNTIL FALSE</lang>
That seems overly complicated. What's wrong with: <lang bbc> REPEAT
INPUT LINE "" q$ space% = INSTR(q$," ") PRINT VAL LEFT$(q$,space%-1) + VAL MID$(q$,space%+1) UNTIL FALSE</lang>
bc
<lang bc>read() + read()</lang>
Befunge
<lang befunge>&&+.@</lang>
Bracmat
filter
is a pattern that checks that input is a non-fractional number not less than -1000 and not greater than 1000. The filter is applied to each input.
<lang bracmat>( out
$ ( put$"Enter two integer numbers between -1000 and 1000:"
& (filter=~/#%:~<-1000:~>1000) & get':(!filter:?a) (!filter:?b) & !a+!b | "Invalid input. Try again" )
);</lang>
Brainf***
<lang brainf***>INPUT AND SUMMATION TODO if first symbol is a minus sign print Qgo awayQ +> initialize sum to one ++[ loop for each input ie twice
[>>,----------[----------------------[-<+>]]<] eat digits until space or newline <[<]>>> >[< until no next digit ---------------- subtract ascii zero minus what we subtracted above [->++++++++++<] add ten timess that to the next digit <[->+<]<[->+<]>> shift sum and loop counter >> ] <---------------- subtract as above from last digit as well [-<<+>>] add to sum <-
] <- subtract original one from sum
OUTPUT [ while a number divided by ten is bigger than zero
[->+<[->+<[->+<[->+<[->+<[->+<[->+<[->+<[->+<[->--------->+<<[->>>+<<<]]]]]]]]]]>>>[-<<<+>>>]<<<] divide by ten >++++++++++++++++++++++++++++++++++++++++++++++++> convert remainder to ascii digit
] <[.<<] print ascii digits</lang>
Brat
<lang brat>numbers = g.split[0,1].map(:to_i) p numbers[0] + numbers[1] #Prints the sum of the input</lang>
Burlesque
<lang burlesque>ps++</lang>
C
<lang c>// Standard input-output streams
- include <stdio.h>
int main() {
int a, b; scanf("%d%d", &a, &b); printf("%d\n", a + b); return 0;
}</lang> <lang c>// Input file: input.txt // Output file: output.txt
- include <stdio.h>
int main() {
freopen("input.txt", "rt", stdin); freopen("output.txt", "wt", stdout); int a, b; scanf("%d%d", &a, &b); printf("%d\n", a + b); return 0;
}</lang> <lang c>
- include <stdio.h>
- include <stdlib.h>
int main(int argc, char **argv) //not sure if argv counts as input stream... certainly it is brought here via input stream. {
printf("%d\n", atoi(*(argv+1)) + atoi(*(argv+2))); return 0;
} </lang>
C#
<lang csharp>using System; using System.Linq;
class Program {
static void Main() { Console.WriteLine(Console.ReadLine().Split().Select(int.Parse).Sum()); }
}</lang> Another way (not recommended since it does not work with more than two numbers): <lang csharp>using System;
class Program
{ static void Main() { start: string input = Console.ReadLine(); int index = input.IndexOf(" "); string num1 = input.Substring(0, index); string num2 = input.Substring(index + 1); int sum = int.Parse(num1) + int.Parse(num2); Console.WriteLine(sum.ToString()); goto start; } }</lang>
C++
<lang cpp>// Standard input-output streams
- include <iostream>
using namespace std; void main() {
int a, b; cin >> a >> b; cout << a + b << endl;
}</lang> <lang cpp>// Input file: input.txt // Output file: output.txt
- include <fstream>
using namespace std; int main() {
ifstream in("input.txt"); ofstream out("output.txt"); int a, b; in >> a >> b; out << a + b << endl; return 0;
}</lang>
Clojure
<lang clojure>(println (+ (Integer/parseInt (read-line)) (Integer/parseInt (read-line)))) 3 4 =>7</lang> <lang clojure>(eval (read-string (str "(+ " (read-line) " )") )) 3 3 6</lang>
COBOL
<lang cobol> IDENTIFICATION DIVISION.
PROGRAM-ID. A-Plus-B.
DATA DIVISION. WORKING-STORAGE SECTION. 01 A PIC S9(5). 01 B PIC S9(5).
01 A-B-Sum PIC S9(5).
PROCEDURE DIVISION. ACCEPT A ACCEPT B
ADD A TO B GIVING A-B-Sum
DISPLAY A-B-Sum
GOBACK .</lang>
CoffeeScript
<lang html4strict><html> <script type="text/javascript" src="http://jashkenas.github.com/coffee-script/extras/coffee-script.js"></script> <script type="text/coffeescript"> a = window.prompt 'enter A number', b = window.prompt 'enter B number', document.getElementById('input').innerHTML = a + ' ' + b sum = parseInt(a) + parseInt(b) document.getElementById('output').innerHTML = sum </script> <body>
</body> </html></lang>
<lang coffeescript> { stdin } = process sum = ( a, b ) -> a + b
display = ( messages... ) -> console.log messages...
parse = ( input ) ->
parseInt x for x in ( x.trim() for x in input.split ' ' ) when x?.length
check = ( numbers... ) ->
return no for x in numbers when isNaN x return no for x in numbers when not ( -1000 < x < 1000 ) yes
prompt = ->
display 'Please enter two integers between -1000 and 1000, separated by a space:' stdin.once 'data', ( data ) -> [ a, b ] = parse data if check a, b display "#{ a } + #{ b } = #{ sum a, b }" else display "Invalid input: #{ a }, #{ b }" do prompt return
- Resume input and set the incoming encoding.
stdin.resume() stdin.setEncoding 'utf8'
- Start the main loop.
do prompt </lang>
Common Lisp
<lang lisp>(write (+ (read) (read)))</lang>
Component Pascal
BlackBox Component Builder <lang oberon2> MODULE AB; IMPORT StdLog, DevCommanders,TextMappers;
PROCEDURE DoAB(x,y: INTEGER); BEGIN
StdLog.Int(x);StdLog.Int(y);StdLog.Int(x + y);StdLog.Ln;
END DoAB;
PROCEDURE Go*; VAR
params: DevCommanders.Par; s: TextMappers.Scanner; p : ARRAY 2 OF INTEGER; current: INTEGER;
BEGIN
current := 0; params := DevCommanders.par; s.ConnectTo(params.text); s.SetPos(params.beg); s.Scan; WHILE(~s.rider.eot) DO IF (s.type = TextMappers.int) THEN p[current] := s.int; INC(current); END; s.Scan; END; IF current = 2 THEN DoAB(p[0],p[1]) END;
END Go;
END AB.
</lang>
Execute: AB.Go 12 23 ~
Output:
12 23 35
D
From Console
<lang d>import std.stdio, std.conv, std.string;
void main() {
string[] r; try r = readln().split(); catch (StdioException e) r = ["10", "20"];
writeln(to!int(r[0]) + to!int(r[1]));
}</lang>
- Output:
30
From File
<lang d>void main() {
import std.stdio, std.file;
immutable ab = "sum_input.txt".slurp!(int, int)("%d %d")[0]; "sum_output.txt".File("w").writeln(ab[0] + ab[1]);
}</lang>
Dart
<lang Dart>import 'dart:io';
// a little helper function that checks if the string only contains // digits and an optional minus sign at the front bool isAnInteger(String str) => str.contains(new RegExp(r'^-?\d+$'));
void main() {
while(true) { String input = stdin.readLineSync(); var chunks = input.split(new RegExp(r'[ ]+')); // split on 1 or more spaces if(!chunks.every(isAnInteger)) { print("not an integer!"); } else if(chunks.length > 2) { print("too many numbers!"); } else if(chunks.length < 2) { print('not enough numbers!'); } else { // parse the strings into integers var nums = chunks.map((String s) => int.parse(s)); if(nums.any((num) => num < -1000 || num > 1000)) { print("between -1000 and 1000 please!"); } else { print(nums.reduce((a, b) => a + b)); } } }
} </lang>
- Output:
1 2 3 3 4 7
dc
<lang dc>? + psz</lang>
The question mark ? reads and executes a line of input. The user must enter a dc program that pushes two numbers to the stack, such as 2 3 or 5 _1. (The user must use underscore _ for negative numbers.)
Déjà Vu
Console
<lang dejavu>0 for k in split !prompt "" " ": + to-num k !print</lang>
Delphi
Console version. <lang delphi>program SUM;
{$APPTYPE CONSOLE}
uses
SysUtils;
var
s1, s2:string;
begin
ReadLn(s1); Readln(s2); Writeln(StrToIntDef(s1, 0) + StrToIntDef(s2,0));
end.</lang>
DMS
<lang DMS>number a = GetNumber( "Please input 'a'", a, a ) // prompts for 'a' number b = GetNumber( "Please input 'b'", b, b ) // prompts for 'b' Result( a + b + "\n" )</lang>
DWScript
Ghetto GUI version <lang delphi>var a := StrToInt(InputBox('A+B', 'Enter 1st number', '0')); var b := StrToInt(InputBox('A+B', 'Enter 2nd number', '0')); ShowMessage('Sum is '+IntToStr(a+b));</lang>
Eiffel
argument(0) contains the path of the executable - thus we start at argument(1) <lang eiffel> class APPLICATION inherit ARGUMENTS create make feature {NONE} -- Initialization make -- Run application. do print(argument(1).to_integer + argument(2).to_integer) end end </lang>
Ela
<lang ela>open console list string read
readn() |> string.split " " |> map readStr |> sum</lang>
Output:
1 2 3 4 5 6 21
Elena
<lang elena>#define system.
- define extensions.
- symbol program =
[
#var A := Integer new. #var B := Integer new. consoleEx readLine:A:B. consoleEx writeLine:(A + B).
]. </lang>
Erlang
<lang erlang>-module(aplusb). -export([start/0]).
start() ->
case io:fread("","~d~d") of eof -> ok; {ok, [A,B]} -> io:format("~w~n",[A+B]), start() end.</lang>
Euler Math Toolbox
<lang Euler Math Toolbox> >s=lineinput("Two numbers seperated by a blank");
Two numbers seperated by a blank? >4 5
>vs=strtokens(s)
4 5
>vs[1]()+vs[2]()
9
</lang>
Euphoria
<lang euphoria>include get.e
function snd(sequence s)
return s[2]
end function
integer a,b
a = snd(get(0)) b = snd(get(0))
printf(1," %d\n",a+b)</lang>
EGL
<lang EGL> package programs;
// basic program // program AplusB type BasicProgram {} function main() try arg1 string = SysLib.getCmdLineArg(1); arg2 string = SysLib.getCmdLineArg(2); int1 int = arg1; int2 int = arg2; sum int = int1 + int2; SysLib.writeStdout("sum1: " + sum); onException(exception AnyException) SysLib.writeStdout("No valid input. Provide 2 integer numbers as arguments to the program."); end end end </lang>
Excel
Take any 3 columns of any row or rows. Let's say A1,B1 and C1 are taken. In C1 type in :
<lang excel> =A1+B1 </lang>
The value of C1 will change as the values of A1 and B1 are changed
<lang>1 2 3 </lang>
F#
<lang fsharp>open System let plus = (fun (a:string) (b:string) -> Console.WriteLine(int(a)+int(b))) (Console.ReadLine()) (Console.ReadLine());;</lang>
Factor
<lang factor>USING: math.parser splitting ;
- a+b ( -- )
readln " " split1 [ string>number ] bi@ + number>string print ;</lang>
( scratchpad ) a+b 2 2 4
FALSE
<lang false>[0[^$$'9>'0@>|~]['0-\10*+]#%]n: {read an integer} n;!n;!+.</lang>
Fantom
<lang fantom>class APlusB {
public static Void main () { echo ("Enter two numbers: ") Str input := Env.cur.in.readLine Int sum := 0 input.split.each |n| { sum += n.toInt } echo (sum) }
}</lang>
FBSL
Using stdin and stdout <lang qbasic>#APPTYPE CONSOLE
DIM %a, %b SCANF("%d%d", @a, @b) PRINT a, "+", b, "=", a + b
PAUSE</lang>
Fish
<lang Fish>i:o:"-"=?v1$68*-v v >~01-0 > >i:o:" "=?v68*-$a*+
>~*i:o:"-"=?v1$68*-v
v >~01-0 > >i:o:d=?v68*-$a*+
>~*+aonao;</lang>
Forth
<lang Forth>pad dup 80 accept evaluate + .</lang>
Fortran
<lang fortran>program a_plus_b
implicit none integer :: a,b read (*, *) a, b write (*, '(i0)') a + b
end program a_plus_b</lang>
Frink
This program handles arbitrarily-large integers, or even floating-point or rational numbers or complex numbers (as long as they're not internally separated with spaces, of course, which are the delimiters for this task.) It can even handle units of measure (with no embedded spaces) such as "3.3meter 2feet" and does the right thing when summing those units. It can handle any number of arbitrary whitespace characters separating the numbers. It also works whether the input is user-interactive, or input comes from stdin or a pipe. (It will bring up a user dialog for input when run in a graphical environment.) <lang frink> sum[eval[split[%r/\s+/, input[""]]]] </lang>
FunL
<lang funl>println( sum(map(int, readLine().split(' +'))) )</lang>
GML
<lang GML>var add, a, b; add = argument0; // get the string with the numbers to add a = real(string_copy(add, 1, string_pos(" ", add))); b = real(string_copy(add, string_pos(" ", add) + 1, string_length(add) - string_pos(" ", add))); return(a + b);</lang>
Go
<lang go>package main
import "fmt"
func main() {
var a, b int fmt.Scan(&a, &b) fmt.Println(a + b)
}</lang>
Golfscript
<lang golfscript>~+</lang>
Gosu
<lang Gosu> uses java.io.InputStreamReader uses java.util.Scanner uses java.lang.System
var scanner = new Scanner( new InputStreamReader( System.in ) ) var a = scanner.nextInt() var b = scanner.nextInt()
print( a + b ) </lang>
Groovy
<lang groovy>def abAdder = {
def reader = new Scanner(System.in) def a = reader.nextInt(); def b = reader.nextInt(); assert (-1000..1000).containsAll([a,b]) : "both numbers must be between -1000 and 1000 (inclusive)" a + b
} abAdder()</lang>
GUISS
We cannot use variables, but we can find the sum of two numbers.Here we add 3 + 2: <lang guiss>Start,Programs,Accessories,Calculator,Button:3,Button:[plus], Button:2,Button:[equals]</lang>
Haskell
<lang haskell>main = getLine >>= print . sum . map read . words</lang>
HicEst
A and B are input via edit controls with spinners limiting inputs to +-1000. <lang HicEst>DLG(Edit=A, DNum, MIn=-1000, MAx=1000, E=B, DN, MI=-1000, MA=1000) WRITE(Messagebox, Name) A, B, "Sum = ", A+B</lang>
Icon and Unicon
<lang icon>procedure main()
numChars := '-'++&digits read() ? { A := (tab(upto(numChars)), integer(tab(many(numChars)))) B := (tab(upto(numChars)), integer(tab(many(numChars)))) } write((\A + \B) | "Bad input")
end</lang>
J
Typically, in J, you would find the sum of two numbers (let us say 2 and 3) by entering both of them on a line with a + sign between them: <lang J> 2+3 5</lang> In the following expression, 1!:1(3) reads a line from STDIN; -.LF drops the line ending character; ". converts the remaining text to a sequence of numbers which are then summed using +/. <lang J>+/". (1!:1(3))-.LF</lang> Here's a little script, called "a+b.ijs": <lang J>#!/Applications/j602/bin/jconsole echo +/". (1!:1(3))-.LF exit </lang> Here is the execution of the script: <lang bash>echo 2 3 | ./a+b.ijs 5</lang>
Java
<lang java>import java.util.*;
public class Sum2 {
public static void main(String[] args) { Scanner in = new Scanner(System.in); // Standard input System.out.println(in.nextInt() + in.nextInt()); // Standard output }
}</lang> Object of class Scanner works slow enough, because of that contestants prefer to avoid its use. Often, longer solution works faster and easily scales to problems. <lang java>import java.io.*; import java.util.*;
public class SumDif {
StreamTokenizer in; PrintWriter out;
public static void main(String[] args) throws IOException { new SumDif().run(); }
private int nextInt() throws IOException { in.nextToken(); return (int)in.nval; }
public void run() throws IOException { in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); // Standard input out = new PrintWriter(new OutputStreamWriter(System.out)); // Standard output solve(); out.flush(); }
private void solve() throws IOException { out.println(nextInt() + nextInt()); }
}</lang> <lang java>import java.io.*;
public class AplusB { public static void main(String[] args) { try { StreamTokenizer in = new StreamTokenizer(new FileReader("input.txt")); in.nextToken(); int a = (int) in.nval; in.nextToken(); int b = (int) in.nval; FileWriter outFile = new FileWriter("output.txt"); outFile.write(Integer.toString(a + b)); outFile.close(); } catch (IOException e) { System.out.println("IO error"); } } }</lang>
JavaScript
ES5
Client side:
<lang html4strict><html> <body>
<script type='text/javascript'> var a = window.prompt('enter A number', ); var b = window.prompt('enter B number', ); document.getElementById('input').innerHTML = a + ' ' + b;
var sum = Number(a) + Number(b); document.getElementById('output').innerHTML = sum; </script> </body> </html></lang>
Server side (with node.js):
<lang javascript>process.openStdin().on (
'data', function (line) { var xs = String(line).match(/^\s*(\d+)\s+(\d+)\s*/) console.log ( xs ? Number(xs[1]) + Number(xs[2]) : 'usage: <number> <number>' ) process.exit() }
)</lang>
$ node io.js 2 3 5 $ node io.js x 3 usage: <integer> <integer>
ES6
Node.js in a terminal: <lang javascript>process.stdin.on("data", buffer => {
console.log( (buffer + "").trim().split(" ").map(Number).reduce((a, v) => a + v, 0) );
}); </lang>
$ node io.js 2 3 5
Julia
Run from the command line: <lang julia>
- A+B
function AB() input = sum(map(int,split(readline(STDIN)," "))) println(input) end AB()</lang>
- Output:
>julia AB.jl 1 1 2
In the next solution, an error is returned if the entry is not constituted from exactly two integers. Any number of spaces can follow an integer. <lang Julia>julia> int(readuntil(STDIN, ' ')) + int(readuntil(STDIN, '\n')) 1 2 3 </lang>
Joy
Console
<lang Joy>get get +.</lang>
File
<lang Joy>"input.txt" include "output.txt" "w" fopen get get + fput pop quit.</lang>
jq
Since the given task is simply to add two numbers, the simplest approach in jq is illustrated by the following transcript: <lang jq>$ jq -s add 3 2 5 </lang> This will work provided the numbers are neither too small nor too large. However, the above program will add **all** the numbers presented on the stream (assuming only numbers are presented). If the task were to add consecutive pairs of numbers, then the approach illustrated in the following transcript can be used, in conjunction with the jq "-s" option:<lang jq> def addpairs:
if length < 2 then empty else (.[0] + .[1]), (.[2:] | addpairs) end;
addpairs</lang> For example, here is a transcript that assumes the program is in a file named AB.jq:<lang jq> $ jq -s -f AB.jq 1 2 3 4 5 6 3 7 11</lang>
L++
<lang lisp>(main
(decl int a) (decl int b) (>> std::cin a b) (prn (+ a b)))</lang>
Lasso
<lang lb>[a + b]</lang>
Lang5
<lang lang5>read read + .
read " " split expand drop + .</lang>
Liberty BASIC
<lang lb>input, n$ print eval(word$(n$,1);" + ";word$(n$,2))</lang>
Lisaac
<lang lisaac>Section Header
+ name := A_PLUS_B
Section Public
- main <- ( (IO.read_integer; IO.last_integer) + (IO.read_integer; IO.last_integer) ).println;</lang>
Logo
<lang logo>show apply "sum readlist</lang>
Lua
<lang Lua>a,b = io.read("*number", "*number") print(a+b)</lang>
Kite
<lang Kite>#!/usr/bin/kite
import "System.file";
in = System.file.stdin; line = in|readline; while(not (line is null)) [
arry = line|split(" "); result = (arry[0])|int + (arry[1])|int; result|print;
line = in|readline;
];</lang> Output: <lang Kite> $ kite a_plus_b.kt <<EOF 5 6 EOF 11 $</lang>
M4
<lang M4> define(`sumstr', `eval(patsubst(`$1',` ',`+'))')
sumstr(1 2) 3</lang>
Maple
<lang maple> convert( scanf( "%d %d" ), '`+`' ); 23 34
57</lang>
Mathematica
Interactive in a notebook <lang Mathematica>Input[] + Input[]</lang>
MATLAB / Octave
<lang MATLAB>function sumOfInputs = APlusB()
inputStream = input('Enter two numbers, separated by a space: ', 's'); numbers = str2num(inputStream); %#ok<ST2NM> if any(numbers < -1000 | numbers > 1000) warning('APlusB:OutOfRange', 'Some numbers are outside the range'); end sumOfInputs = sum(numbers);
end</lang>
Maude
Built-in
<lang Maude> red 3 + 4 . </lang>
With restrictions
<lang Maude> fmod ADD is
protecting INT .
op undefined : -> Int . op _add_ : Int Int -> Int [assoc comm] .
vars A B : Int .
eq A add B = if (A < -1000 or B < -1000) or (A > 1000 or B > 1000) then undefined else A + B fi .
endfm </lang>
Mercury
<lang>:- module a_plus_b.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module int, list, string.
main(!IO) :-
io.read_line_as_string(Result, !IO), ( if Result = ok(Line), [AStr, BStr] = string.words(Line), string.to_int(AStr, A), string.to_int(BStr, B) then io.format("%d\n", [i(A + B)], !IO) else true ).</lang>
Maxima
<lang>in_stream: openr("/dev/stdin"); unless (line: readline(in_stream), line=false) do (
q: map('parse_string, split(line, " ")), print(q[1]+q[2]) );
close(in_stream); </lang>
mIRC Scripting Language
<lang mirc>alias a+b {
echo -ag $calc($1 + $2)
}</lang>
МК-61/52
С/П + С/П
ML/I
The two numbers are read from 'standard input' or its equivalent. <lang ML/I>MCSKIP "WITH" NL "" A+B "" assumes macros on input stream 1, terminal on stream 2 MCSKIP MT,<> MCINS %. MCDEF SL SPACES NL AS <MCSET T1=%A1. MCSET T2=%A2. %T1+T2. MCSET S10=0 > MCSKIP SL WITH * MCSET S1=1
- MCSET S10=2</lang>
Modula-2
<lang modula2>MODULE ab;
IMPORT InOut;
VAR A, B : INTEGER;
BEGIN
InOut.ReadInt (A); InOut.ReadInt (B); InOut.WriteInt (A + B, 8); InOut.WriteLn
END ab.</lang>
MoonScript
<lang moonscript>a,b = io.read '*number','*number' print a + b</lang>
MUMPS
<lang MUMPS>ANB
NEW A,B,T,S READ !,"Input two integers between -1000 and 1000, separated by a space: ",S SET A=$PIECE(S," ",1),B=$PIECE(S," ",2) SET T=(A>=-1000)&(A<=1000)&(B>=-1000)&(B<=1000)&(A\1=A)&(B\1=B) IF T WRITE !,(A+B) IF 'T WRITE !,"Bad input" QUIT</lang>
Nemerle
<lang Nemerle>using System; using System.Console; using System.Linq;
module AplusB {
Main() : void { WriteLine(ReadLine().Split().Select(int.Parse).Sum()); }
}</lang>
NetRexx
<lang NetRexx>/* NetRexx */
options replace format comments java crossref savelog symbols binary
parse ask a b . say a '+' b '=' a + b</lang>
newLISP
<lang newLISP>(println (apply + (map int (parse (read-line)))))</lang>
Nimrod
A+B: <lang nimrod>import strutils, os
echo parseInt(paramStr(1)) + parseInt(paramStr(2))</lang> Arbitrary number of arguments: <lang nimrod>import strutils, os var sum = 0 for i in countup(1, paramCount()):
sum = sum + parseInt(paramStr(i))
echo sum</lang>
another:
<lang nimrod>from strutils import parseFloat, formatFloat, ffDecimal
proc aplusb(a,b: float): float =
return a + b
proc getnumber(): float =
try: parseFloat(readLine(stdin)) except EInvalidValue: echo("Please enter a number: ") getnumber()
echo("First number please: ") let first: float = getnumber()
echo("Second number please: ") let second: float = getnumber()
echo("Result: " & formatFloat(aplusb(first, second), ffDecimal, 2))</lang>
Objeck
<lang objeck>bundle Default {
class Vander { function : Main(args : String[]) ~ Nil { values := IO.Console->ReadString()->Split(" "); if(values->Size() = 2) { (values[0]->Trim()->ToInt() + values[1]->Trim()->ToInt())->PrintLine(); }; } }
}</lang>
Oberon-2
<lang oberon2>MODULE ab;
IMPORT In, Out;
VAR A, B : INTEGER;
BEGIN
In.Int (A); In.Int (B); Out.Int (A + B, 8); Out.Ln
END ab.</lang> Producing
12 34 46
OCaml
<lang ocaml>Scanf.scanf "%d %d" (fun a b -> Printf.printf "%d\n" (a + b))</lang>
OpenEdge/Progress
<lang progress>DEFINE VARIABLE a AS INTEGER NO-UNDO FORMAT "->>>9". DEFINE VARIABLE b AS INTEGER NO-UNDO FORMAT "->>>9".
IF SESSION:BATCH THEN DO:
INPUT FROM "input.txt". IMPORT a b. INPUT CLOSE.
END. ELSE
UPDATE a b.
MESSAGE a + b VIEW-AS ALERT-BOX</lang>
Openscad
<lang openscad>a = 5 + 4; echo (a);</lang>
Oxygene
<lang oxygene> // Sum 2 integers read fron standard input // // Nigel Galloway - April 16th., 2012 // namespace aplusb;
interface
uses System.Text.RegularExpressions.*;
type
aplusb = class public class method Main; end;
implementation
class method aplusb.Main; var
gc: GroupCollection; m : Match;
begin
m := new Regex('^\s*(?<a>-?[1-9]\d{0,2}|0|-?1000)\s+(?-?[1-9]\d{0,2}|0|-?1000)\s*$').Match(Console.ReadLine()); if m.Success then begin gc := m.Groups; Console.WriteLine("{0} + {1} = {2}", gc['a'].Value, gc['b'].Value, Integer.Parse(gc['a'].Value) + Integer.Parse(gc['b'].Value)); end else Console.WriteLine("Invalid Input");
end;
end. </lang> Produces:
>aplusb 23 -99 23 + -99 = -76
Oz
<lang oz>declare
class TextFile from Open.file Open.text end
StdIn = {New TextFile init(name:stdin)}
fun {ReadInt} {String.toInt {StdIn getS($)}} end
in
{Show {ReadInt}+{ReadInt}}</lang>
PARI/GP
User input: <lang parigp>input()+input()</lang> File input: <lang parigp>read("file1")+read("file2")</lang>
Pascal
<lang pascal>var
a, b: integer;
begin
readln(a, b); writeln(a + b);
end.</lang> Same with input from file input.txt and output from file output.txt. <lang pascal>var
a, b: integer;
begin
reset(input, 'input.txt'); rewrite(output, 'output.txt'); readln(a, b); writeln(a + b); close(input); close(output);
end.</lang>
Perl
<lang Perl>my ($a,$b) = split(' ', scalar(<STDIN>)); print "$a $b " . ($a + $b) . "\n";</lang>
Perl 6
<lang perl6>say [+] .words for lines</lang>
PHP
<lang php>fscanf(STDIN, "%d %d\n", $a, $b); //Reads 2 numbers from STDIN echo ($a + $b) . "\n";</lang> <lang php>$in = fopen("input.dat", "r"); fscanf($in, "%d %d\n", $a, $b); //Reads 2 numbers from file $in fclose($in);
$out = fopen("output.dat", "w"); fwrite($out, ($a + $b) . "\n"); fclose($out);</lang>
PicoLisp
<lang PicoLisp>(+ (read) (read)) 3 4 -> 7</lang>
Piet
The code is fairly straightforward. The individual commands are as follows:
<lang text>in(num)
in(num)
add
out(num)</lang>
Pike
<lang Pike>string line = Stdio.stdin->gets(); sscanf(line, "%d %d", int a, int b); write(a+b +"\n");</lang>
PL/I
<lang PL/I>get (a, b); put (a+b);</lang>
PostScript
<lang postscript>(%stdin) (r) file % get stdin dup token pop % read A exch token pop % read B add =</lang>
PowerShell
<lang powershell>$a,$b = -split "$input" [int]$a + [int]$b</lang> This solution does not work interactively, while the following only works interactively: <lang powershell>$a,$b = -split (Read-Host) [int]$a + [int]$b</lang>
ProDOS
With the math module: <lang ProDOS>editvar /newvar /value=a /title=Enter an integer: editvar /newvar /value=b /title=Enter another integer: editvar /newvar /value=c do add -a-,-b-=-c- printline -c- </lang> Without the math module: <lang ProDOS>editvar /newvar /value=a /title=Enter an integer: editvar /newvar /value=b /title=Enter another integer: editvar /newvar /value=c=-a-+-b- printline -c- </lang>
Prolog
<lang Prolog>plus :-
read_line_to_codes(user_input,X), atom_codes(A, X), atomic_list_concat(L, ' ', A), maplist(atom_number, L, LN), sumlist(LN, N), write(N).</lang>
output : <lang Prolog>?- plus. |: 4 5 9 true.</lang>
Pure
<lang pure>using system; printf "%d\n" (x+y) when x,y = scanf "%d %d" end;</lang>
PureBasic
Console
<lang PureBasic>x$=Input() a=Val(StringField(x$,1," ")) b=Val(StringField(x$,2," ")) PrintN(str(a+b))</lang>
File
<lang PureBasic>If ReadFile(0,"in.txt")
x$=ReadString(0) a=Val(StringField(x$,1," ")) b=Val(StringField(x$,2," ")) If OpenFile(1,"out.txt") WriteString(1,str(a+b)) CloseFile(1) EndIf CloseFile(0)
EndIf </lang>
Python
Console
<lang python>try: raw_input except: raw_input = input
print(sum(int(x) for x in raw_input().split()))</lang>
File
For Python 2.X and 3.X taking input from stdin stream which can be redirected to be file input under Unix <lang python>import sys
for line in sys.stdin:
print(sum(int(i) for i in line.split()))</lang>
R
<lang r>sum(scan("", numeric(0), 2))</lang>
Racket
<lang racket>
- lang racket
(+ (read) (read)) </lang>
Or, with additional error checking: <lang racket>
- lang racket
(define a (read)) (unless (number? a) (error 'a+b "number" a)) (define b (read)) (unless (number? b) (error 'a+b "number" b)) (displayln (+ a b)) </lang>
REBOL
<lang rebol>forever [x: load input print x/1 + x/2]</lang> Sample output:
1 2 3 2 2 4 3 2 5
Retro
<lang Retro>: try ( "-n ) getToken toNumber getToken toNumber + putn ;</lang> <lang Retro>try 1 2</lang>
REXX
version 1, unnormalized
The numbers can be any valid REXX number (integer, fixed point decimal, floating point (with exponential notation, ...). <lang rexx>parse pull a b say a+b</lang>
version 2, normalizied
If the user entered 4.00000 and wanted to add 5 to that, and expects 9,
then the output needs to be normalized before displaying the result.
Normally, REXX will keep the greatest precision in the results;
adding 4.00000 and 5 will yield 9.00000
<lang rexx>parse pull a b
say (a+b)/1 /*dividing by 1 normalizes the REXX number.*/</lang>
version 3, extended precision
Using the numeric digits statement allows more digits to be used, the default is 9. <lang rexx>numeric digits 300 parse pull a b z=(a+b)/1 say z</lang>
version 4, multiple numbers
This REXX program version adds all the numbers entered (not just two). <lang rexx>numeric digits 1000 /*just in case the user gets ka-razy. */ say 'enter some numbers to be summed:' parse pull y many=words(y) sum=0
do j=1 for many sum=sum+word(y,j) end
say 'sum of' many "numbers = " sum/1</lang>
Ruby
<lang ruby>puts gets.split.map{|x| x.to_i}.inject{|sum, x| sum + x}</lang>
<lang ruby>puts gets.split.map(&:to_i).inject(&:+)</lang>
Run BASIC
<lang runbasic>input, x$ print val(word$(x$,1)) + val(word$(x$,2))</lang>
Rust
<lang Rust>// -*- rust v0.9 -*- use std::os;
fn main() {
let args : ~[~str] = os::args();
let mut values = 0;
for i in args.iter(){ match from_str::<int>(i.to_str()) { Some(valid_int) => values += valid_int, None => () } } println(values.to_str()); } </lang>
Scala
<lang scala>println(readLine().split(" ").map(_.toInt).sum)</lang>
This will work if the input is exactly as specified, with no extra whitespace. A slightly more robust version:
<lang scala>val s = new java.util.Scanner(System.in) val sum = s.nextInt() + s.nextInt() println(sum)</lang>
Scheme
<lang scheme>(display (+ (read) (read)))</lang>
sed
Sed is for string processing and has no facility for manipulating numbers as numeric values. However, being Turing complete, sed can be coerced into performing mathematics. <lang sed>: Loop
- All done
/^-*00* /s/// / -*00*$/s/// t
- Negative Check
/^\(-*\)[0-9].* \1[0-9]/!b Negative
- Create magic lookup table
s/\(.[0-9]*\) \(.[0-9]*\)/\1;987654321000009999000999009909 \2;012345678999990000999000990090/ s/ \(-\)*\(9*;\)/ \10\2/
- Decrement 1st number
s/\([^0]\)\(0*\);[^0]*\1\(.\).*\2\(9*\).* \(.*\)/\3\4 \5/
- Increment 2nd
s/\([^9]\)\(9*\);[^9]*\1\(.\).*\2\(0*\).*/\3\4/ t Loop
- Negative
- Create magic lookup table
s/\(.[0-9]*\) \(.[0-9]*\)/\1;987654321000009999000999009909 \2;987654321000009999000999009909/
- Decrement 1st number
s/\([^0]\)\(0*\);[^0]*\1\(.\).*\2\(9*\).* \(.*\)/\3\4 \5/
- Decrement 2nd
s/\([^0]\)\(0*\);[^0]*\1\(.\).*\2\(9*\).*/\3\4/ t Loop</lang>
Another method, based off of this StackExchange answer: <lang sed>#!/bin/sed -f
- Add a marker in front of each digit, for tracking tens, hundreds, etc.
s/[0-9]/<&/g
- Convert numbers to, in essence, tally marks
s/0//g; s/1/|/g; s/2/||/g; s/3/|||/g; s/4/||||/g; s/5/|||||/g s/6/||||||/g; s/7/|||||||/g; s/8/||||||||/g; s/9/|||||||||/g
- Multiply by ten for each digit from the back they were.
- tens
s/|</<||||||||||/g t tens
- We don't want the digit markers any more
s/<//g
- Negative minus negative is the negation of their absolute values.
s/^-\(|*\) *-/-\1/
- Negative plus positive equals positive plus negative, and we want the negative at the back.
s/^-\(|*\) \+\(|*\)$/\2-\1/
- Get rid of any space between the numbers
s/ //g
- A tally on each side can be canceled.
- minus
s/|-|/-/ t minus s/-$//
- Convert back to digits
- back
s/||||||||||/</g s/<\([0-9]*\)$/<0\1/g s/|||||||||/9/g; s/|||||||||/9/g; s/||||||||/8/g; s/|||||||/7/g; s/||||||/6/g; s/|||||/5/g; s/||||/4/g; s/|||/3/g; s/||/2/g; s/|/1/g; s/</|/g t back s/^$/0/</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
const proc: main is func
local var integer: a is 0; var integer: b is 0; begin read(a); read(b); writeln(a + b); end func;</lang>
Shiny
<lang shiny>if (io.line 'stdin').match ~(\d+)\s+(\d+)~
say "$a $b %(a+b)d"
end</lang>
Sidef
Works with both positive and negative integers. <lang ruby>say STDIN.get.words.map{.toInt}.sum;</lang>
SNOBOL4
Simple-minded solution (literally "two somethings separated by space") <lang snobol> input break(" ") . a " " rem . b output = a + b end</lang> "Integer aware" solution: <lang snobol> nums = "0123456789" input span(nums) . a break(nums) span(nums) . b output = a + b end</lang>
Smalltalk
Most Smalltalk implementations do not have the notion of a standard input stream, since it has always been a GUI based programming environment. I've included test methods to demonstrate one way to create an input stream with two integers can be created. Opening a text file would be another. <lang smalltalk>'From Squeak3.7 of 4 September 2004 [latest update: #5989] on 8 August 2011 at 3:50:55 pm'! Object subclass: #ABTask instanceVariableNames: classVariableNames: poolDictionaries: category: 'rosettacode'!
"-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- "!
ABTask class instanceVariableNames: !
!ABTask class methodsFor: 'demo'! parseInteger: inputStream ^ Integer readFrom: inputStream skipSeparators! !
!ABTask class methodsFor: 'demo'! sum: inputStream ^ (self parseInteger: inputStream) + (self parseInteger: inputStream)! !
!ABTask class methodsFor: 'demo'! test2Plus2 ^ self sum: (ReadStream on: '2 2')! !
!ABTask class methodsFor: 'demo'! test3Plus2 ^ self sum: (ReadStream on: '3 2')! !</lang>
but all have a stream hierarchy, so the task could be restated to pass input and output as stream arguments:
<lang smalltalk>|task| task := [:inStream :outStream |
|processLine|
processLine := [ |a b| a := Integer readFrom: inStream. b := Integer readFrom: inStream. "is validation part of the task?" self assert:( a between:-1000 and: 1000). self assert:( b between:-1000 and: 1000). outStream print (a+b); cr. ].
[ inStream atEnd ] whileFalse:processLine.
].
task value: ( 'dataIn.txt' asFilename readStream) value:Transcript.</lang> or: <lang smalltalk>task value: Stdin value: Stdout.</lang>
SPARK
<lang Ada>-- By Jacob Sparre Andersen -- Validates with SPARK GPL 2010's Examiner/Simplifier
with SPARK_IO; --# inherit SPARK_IO;
--# main_program; procedure A_Plus_B --# global in out SPARK_IO.Inputs, SPARK_IO.Outputs; --# derives SPARK_IO.Inputs from SPARK_IO.Inputs & --# SPARK_IO.Outputs from SPARK_IO.Inputs, SPARK_IO.Outputs; is
subtype Small_Integers is Integer range -1_000 .. +1_000; A, B : Integer; A_OK, B_OK : Boolean;
begin
SPARK_IO.Get_Integer (File => SPARK_IO.Standard_Input, Item => A, Width => 0, Read => A_OK); A_OK := A_OK and A in Small_Integers; SPARK_IO.Get_Integer (File => SPARK_IO.Standard_Input, Item => B, Width => 0, Read => B_OK); B_OK := B_OK and B in Small_Integers; if A_OK and B_OK then SPARK_IO.Put_Integer (File => SPARK_IO.Standard_Output, Item => A + B, Width => 4, Base => 10); else SPARK_IO.Put_Line (File => SPARK_IO.Standard_Output, Item => "Input data does not match specification.", Stop => 0); end if;
end A_Plus_B;</lang>
SQL
<lang sql>select A+B</lang> Example: <lang sql>select 2+3</lang> This should produce a result set containing the value 5.
Note however that declaration of variables is outside the scope of the ANSI SQL standards, unless by variables you mean tables (which would complicate the example considerably).
Tcl
<lang tcl>scan [gets stdin] "%d %d" x y puts [expr {$x + $y}]</lang> Alternatively: <lang tcl>puts [tcl::mathop::+ {*}[gets stdin]]</lang> To/from a file: <lang tcl>set in [open "input.txt"] set out [open "output.txt" w] scan [gets $in] "%d %d" x y puts $out [expr {$x + $y}] close $in close $out</lang>
TI-83 BASIC , TI-89 BASIC
<lang ti83b>:Prompt A,B
- Disp A+B</lang>
Alternate TI-89 BASIC version (returns the result on the Home screen): <lang ti89b>:aplusb(a,b)
- a+b</lang>
TorqueScript
Since torque is not compatible with standard input, I will show the closest to that. It's a function that takes a single string input, that will contain the 2 numbers. <lang Torque>Function aPlusB(%input) {
return getWord(%input, 0) + getWord(%input, 1);
}</lang>
TUSCRIPT
<lang tuscript>$$ MODE TUSCRIPT SET input="1 2" SET input=SPLIT(input,": :") SET input=JOIN (input) SET output=SUM(input)</lang>
TXR
$ txr -p '(+ (read) (read))' 1.2 2.3 3.5
UNIX Shell
<lang sh>#!/bin/sh read a b || exit echo `expr "$a" + "$b"`</lang>
Script "a+b.sh": <lang bash>#!/bin/bash read a b || exit echo $(( a + b ))</lang> Output: <lang bash>echo 2 3 | ksh a+b.sh 5</lang>
C Shell
<lang csh>set line=$< set input=($line) @ sum = $input[1] + $input[2] echo $sum</lang>
Ursala
Using standard input and output streams: <lang Ursala>#import std
- import int
- executable&
main = %zP+ sum:-0+ %zp*FiNCS+ sep` @L</lang> Overwriting a text file named as a command line parameter: <lang Ursala>#import std
- import int
- executable -[parameterized]-
main = ~command.files.&h; <.contents:= %zP+ sum:-0+ %zp*FiNCS+ sep` @L+ ~contents></lang>
Creating a new file named after the input file with suffix .out
:
<lang Ursala>#import std
- import int
- executable -[parameterized]-
main =
~command.files.&h; ~&iNC+ file$[
contents: %zP+ sum:-0+ %zp*FiNCS+ sep` @L+ ~contents, path: ~path; ^|C\~& ~=`.-~; ^|T/~& '.out'!]</lang>
Vala
Read from stdin while program running: <lang vala>stdout.printf("Please enter int value for A\n"); int a = int.parse(stdin.read_line()); stdout.printf("Please enter int value for B\n"); int b = int.parse(stdin.read_line());
stdout.printf("A+B = %d\n", a+b);</lang>
VBA
<lang VBA>Sub Rosetta_AB() Dim stEval As String stEval = InputBox("Enter two numbers, separated only by a space", "Rosetta Code", "2 2") MsgBox "You entered " & stEval & vbCr & vbCr & _
"VBA converted this input to " & Replace(stEval, " ", "+") & vbCr & vbCr & _ "And evaluated the result as " & Evaluate(Replace(stEval, " ", "+")), vbInformation + vbOKOnly, "XLSM"
End Sub</lang>
VBScript
<lang vb>Option Explicit Dim a, b Select Case WScript.Arguments.Count Case 0 'No arguments, prompt for them. WScript.Echo "Enter values for a and b" a = WScript.Stdin.ReadLine if Instr(a, " ") > 0 then 'If two variables were passed b = Split(a)(1) a = Split(a)(0) else WScript.Echo "Enter value for b" b = WScript.Stdin.ReadLine end if Case 1 'One argument, assume it's an input file, e.g. "in.txt" Dim FSO : Set FSO = CreateObject("Scripting.FileSystemObject") With FSO.OpenTextFile(WScript.Arguments(0), 1) a = .ReadLine b = Split(a)(1) a = Split(a)(0) .Close End With Case 2 'Two arguments, assume they are values a = WScript.Arguments(0) b = WScript.Arguments(1) End Select 'At this point, a and b are strings as entered, make them numbers a = CInt(a) b = CInt(b)
'Write the sum Wscript.Echo a + b if 1 = WScript.Arguments.Count then With FSO.CreateTextFile("out.txt") .WriteLine a + b .Close End With end if</lang>
VHDL
<lang VHDL>LIBRARY std; USE std.TEXTIO.all;
entity test is
end entity test;
architecture beh of test is
begin
process variable line_in, line_out : line; variable a,b : integer; begin readline(INPUT, line_in); read(line_in, a); read(line_in, b); write(line_out, a+b); writeline(OUTPUT, line_out); wait; -- needed to stop the execution end process;
end architecture beh;</lang>
Visual Basic .NET
<lang vbnet>Module Module1
Sub Main() Dim s() As String = Nothing
s = Console.ReadLine().Split(" "c) Console.WriteLine(CInt(s(0)) + CInt(s(1))) End Sub
End Module</lang>
Whitespace
<lang whitespace>
</lang>
X86 Assembly
<lang asm>section .text global _start
_print: mov ebx, 1 mov eax, 4 int 0x80 ret
_get_input: mov edx, 4 mov ebx, 0 mov eax, 3 int 0x80 ret
_start: mov edx, in_val_len mov ecx, in_val_msg call _print mov ecx, a call _get_input ;make 'a' an actual number rather than a char. sub dword [a], 0x30 mov edx, in_val_len mov ecx, in_val_msg call _print mov ecx, b call _get_input ;calc real number for 'b' sub dword [b], 0x30 mov eax, dword [a] mov ebx, dword [b] add eax, ebx ;get the character for our sum. add eax, 0x30 mov dword [sum], eax mov edx, out_val_len mov ecx, out_val_msg call _print mov [sum+1], dword 0xa mov edx, 4 mov ecx, sum call _print push 0x1 mov eax, 1 push eax int 0x80 ret
section .data in_val_msg db "Please input an integer:",0 in_val_len equ $-in_val_msg out_val_msg db "The sum of a+b is: ",0 out_val_len equ $-out_val_msg
section .bss a resd 1 b resd 1 sum resd 1</lang> This will not work on numbers over 0(from 1 to 0). This is due to the fact, numbers higher than 0(10,11, etc) are in fact strings when taken from the keyboard. A much longer conversion code is required to loop through and treat each number in the string as separate numbers. For example, The number '10' would have to be treated as a 1 and a 0.
XPL0
<lang XPL0>include c:\cxpl\codes; int A, B; [A:= IntIn(0);
B:= IntIn(0); IntOut(0, A+B); CrLf(0);
]</lang>
Yorick
<lang yorick>a = b = 0; read, a, b; write, a + b;</lang>
ZED
Source -> http://ideone.com/WLtEfe Compiled -> http://ideone.com/fMt6ST <lang zed>(A+B) comment:
- true
(+) (read) (read)
(+) one two comment:
- true
(003) "+" one two
(read) comment:
- true
(001) "read"</lang>
zkl
<lang zkl>do(2){ask("A B: ").split(" ").filter().sum().println()}</lang>
A B: 123 567 690 A B: -4 6 2
This actually works for any number of integers
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