Sum of Array: Difference between revisions

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[ ( array size -- sum )
[ ( array size -- sum )
>r 0 r> [ over i swap get-element + ] iterate nip ] is sum-array
>r 0 r> 0 [ over i swap array.get + ] countedLoop nip ] is sum-array


==[[UNIX Shell]]==
==[[UNIX Shell]]==

Revision as of 11:06, 11 September 2007

Task
Sum of Array
You are encouraged to solve this task according to the task description, using any language you may know.

Compute the sum of the elements of an Array

4D

ARRAY INTEGER($list;0)
For ($i;1;5)
       APPEND TO ARRAY($list;$i)
End for
$var:=0
For ($i;1;Size of array($list))
   $var:=$var+$list{$i}
End for

Ada

Int_Array : array(1..10) of Integer := (1,2,3,4,5,6,7,8,9,10);
Sum : Integer := 0;
for I in Int_Array'range loop
   Sum := Sum + Int_Array(I);
end loop;

AppleScript

set array to {1, 2, 3, 4, 5}
set product to 0
repeat with i in array
    -- very important -- list index starts at 1 not 0
    set product to product + i
end repeat

BASIC

 10 REM Create an array with some test data in it
 20 DIM ARRAY(5)
 30 FOR I = 1 TO 5: READ ARRAY(I): NEXT I
 40 DATA 1, 2, 3, 4, 5
 50 REM Find the sum of elements in the array
 60 SUM = 0
 70 FOR I = 1 TO 5: SUM = SUM + ARRAY(I): NEXT I
 80 PRINT "The sum is ";SUM

C++

Compiler: gcc 4.0.2

int
main( int argc, char* argv[] )
{
 int list[] = { 1, 2, 3 } ;
 int sum = 0 ;
 for( int i = 0 ; i < 3 ; i++ )
 {
  sum += list[i];
 }
}


Alternate

#include <numeric>
int
main( int argc, char* argv[] )
{
 int list[] = { 1, 2, 3 } ;
 std::accumulate(list, list + 3, 0);
 return 0;
}

template alternative

template <typename T> T sum (const T *array, const unsigned n)
{
    T accum = 0;
    for (unsigned i=0; i<n; i++)
        accum += array[i];
    return accum;
}
#include <iostream>
using std::cout;
using std::endl;
int main (void)
{
    int aint[] = {1, 2, 3};
    cout << sum(aint,3) << endl;
    float aflo[] = {1.1, 2.02, 3.003, 4.0004};
    cout << sum(aflo,4) << endl;
    return 0;
}

C#

 int value = 0;
 int[] arg = { 1,2,3,4,5 };
 int arg_length = arg.Length;
 for( int i = 0; i < arg_length; i++ )
    value += arg[i];


Alternate

 int sum = 0;
 int[] arg = { 1, 2, 3, 4, 5 };
 foreach (int value in arg) sum += value;

Clean

array = {1, 2, 3, 4, 5}
Start = sum [x \\ x <-: array]

ColdFusion

 <cfset myArray = listToArray("1,2,3,4,5")>
 #arraySum(myArray)#

Common Lisp

(defparameter *data* #(1 2 3 4 5))
(reduce #'+ *data*)

Delphi

[[Category:Delphi]

Compiler: All

 var
   Ints   : array[1..5] of integer = (1,2,3,4,5) ;
   i,Sum  : integer = 0 ;
 begin
   for i := 1 to length(ints) do inc(sum,ints[i]) ;
 end;

E

pragma.enable("accumulator")
accum 0 for x in [1,2,3,4,5] { _ + x }

Erlang

Using the standard libraries:

% create the list:
L = lists:seq(1, 10).
% and compute its sum:
S = lists:sum(L).

Or defining our own versions:

-module(list_sum).
-export([sum_rec/1, sum_tail/1]).
% recursive definition:
sum_rec([]) ->
    0;
sum_rec([Head|Tail]) ->
    Head + sum_rec(Tail).
% tail-recursive definition:
sum_tail(L) ->
    sum_tail(L, 0).
sum_tail([], Acc) ->
    Acc;
sum_tail([Head|Tail], Acc) ->
    sum_tail(Tail, Head + Acc).

Forth

 : sum ( addr cnt -- n )
   0 -rot
   cells bounds do i @ + cell +loop ;


FreeBASIC

 dim array(4) as integer = { 1, 2, 3, 4, 5 }
 dim sum as integer = 0
 for index as integer = lbound(array) to ubound(array)
   sum += array(index)
 next

Haskell

 let values = [1..10]
 sum values          -- the easy way
 foldl (+) 0 values  -- the hard way

IDL

 result = total(array)

Java

 int value = 0;
 int[] arg = new int[] { 1,2,3,4,5 };
 for (int i: arg)
   value += i;

JavaScript

var array = [1, 2, 3, 4, 5];
var sum = 0;
for(var i in array)
  sum += array[i];

Perl

Interpeter: Perl

my $var;
my @list = (1, 2, 3);
$var += $_ for (@list);

Alternate

Libraries: List::Util

use List::Util 'sum';
my @list = (1, 2, 3);
my $var = sum @list;

Alternate

# TMTOWTDI

my $acc = 0;
my @list = qw(1 2 3)
map { $acc += $_ } @list

PHP

 $list = array(1,2,3,4,5,6,7,8,9);
 echo array_sum($list);

Pop11

Simple loop:

lvars i, sum = 0, ar = {1 2 3 4 5 6 7 8 9};
for i from 1 to length(ar) do
    ar(i) + sum -> sum;
endfor;

One can alternativly use second order iterator:

lvars sum = 0, ar = {1 2 3 4 5 6 7 8 9};
appdata(ar, procedure(x); x + sum -> sum; endprocedure);

Prolog

sum([],0).
sum([H|T],X) :- sum(T,Y), X is H + Y.

test

:- sum([1,2,3,4,5,6,7,8,9],X).
X =45;

Python

Interpeter: Python 2.5

 total = sum([1, 2, 3, 4, 5, 6, 7, 8, 9])

Ruby

 ary = [1,2,3,4,5]
 sum = ary.inject{|currentSum,element|currentSum+element}
 # => 15

Scala

   val array = Array(1,2,3,4,5)
   val sum = array.foldLeft(0)(_ + _)

This is a shortcut for

 val sum = array.foldLeft(0){(currentSum, element) => currentSum + element}

Seed7

const func integer: sumArray (in array integer: valueArray) is func
  result
    var integer: sum is 0;
  local
    var integer: value is 0;
  begin
    for value range valueArray do
      sum +:= value;
    end for;
  end func;

Call this function with:

writeln(sumArray([](1, 2, 3, 4, 5)));

Standard ML

 val array = [1,2,3,4,5];
 foldl op+ 0 array;

Tcl

Assuming the values are in a list named listname:

 set result [expr [join $listname +]]

Toka

[ ( array size -- sum )
  >r 0 r> 0 [ over i swap array.get + ] countedLoop nip ] is sum-array

UNIX Shell

Interpreter: NetBSD 3.0's ash

From an internal variable, $IFS delimited:

 sum=0
 list="1 2 3"
 for n in $list
 do sum="$(($sum + $n))"
 done
 echo $sum

From the argument list (ARGV):

 sum=0
 for n
 do sum="$(($sum + $n))"
 done
 echo $sum

From STDIN, one integer per line:

 sum=0
 while read n
 do sum="$(($sum + $n))"
 done
 echo $sum

Interpreter: GNU bash, version 3.2.0(1)-release (i386-unknown-freebsd6.1)

From variable:

 LIST='20 20 2';
 SUM=0;
 for i in $LIST; do
   SUM=$[$SUM + $i];
 done;
 echo $SUM