Sum of Array: Difference between revisions

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#REDIRECT [[Sum and product of array]]
{{task}}
Compute the sum of the elements of an Array

=={{header|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

=={{header|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;

=={{header|AppleScript}}==
set array to {1, 2, 3, 4, 5}
set product to 0
repeat with i in array
set product to product + i
end repeat

=={{header|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

=={{header|C++}}==

'''Compiler:''' [[GNU Compiler Collection|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;
}

=={{header|C sharp|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;

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

=={{header|ColdFusion}}==
<cfset myArray = listToArray("1,2,3,4,5")>
#arraySum(myArray)#

=={{header|Common Lisp}}==
(reduce #'+ #(1 2 3 4 5))

=={{header|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;

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

=={{header|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).

=={{header|Forth}}==
: sum ( addr cnt -- n )
0 -rot
cells bounds do i @ + cell +loop ;


=={{header|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

=={{header|Groovy}}==
[1,2,3,4,5].sum()

[1,2,3,4,5].inject(0) { sum, val -> sum + val }

=={{header|Haskell}}==

For lists (''sum'' is already defined in the Prelude):

values = [1..10] :: [Int]
answer = sum values -- the easy way
answer' = foldl (+) 0 values -- the hard way

To do the same for an array, just convert it lazily to a list:

import Data.Array
values = listArray (1,10) [1..10]
answer = sum $ elems $ values

==[[IDL]]==
[[Category:IDL]]

result = total(array)

=={{header|Java}}==

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

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

=={{header|MAXScript}}==
arr = #(1, 2, 3, 4, 5)
result = 0
for i in arr do result += i

=={{header|OCaml}}==
let a = [| 1; 2; 3; 4; 5 |] in
Array.fold_left (+) 0 a

=={{header|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

=={{header|PHP}}==
$list = array(1,2,3,4,5,6,7,8,9);
echo array_sum($list);

=={{header|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);

=={{header|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;

=={{header|Python}}==
'''Interpeter:''' [[Python]] 2.5
total = sum([1, 2, 3, 4, 5, 6, 7, 8, 9])

=={{header|Raven}}==
0 [ 1 2 3 ] each +

=={{header|Ruby}}==
arr = [1,2,3,4,5]
sum = arr.inject { |sum, item| sum + item }

=={{header|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}

=={{header|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)));

=={{header|Smalltalk}}==
#(1 2 3 4 5) inject: 0 into: [:sum :number | sum + number]

=={{header|Standard ML}}==
val array = [1,2,3,4,5];
foldl op+ 0 array;

=={{header|Tcl}}==
Assuming the values are in a list named <tt>listname</tt>:

set result [expr [join $listname +]]

=={{header|Toka}}==
[ ( array size -- sum )
>r 0 r> 0 [ over i swap array.get + ] countedLoop nip ] is sum-array

=={{header|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

Revision as of 23:39, 3 December 2007

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