Calculate the sum over a container

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Revision as of 17:55, 11 January 2009 by (Talk)

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This code demonstrates how to calculate the sum of a container (most often an array or similar structure).

See also


Algol 68

The following function computes the sum of the elements of the array passed to it. The type of the array elements is called NUM, which can be defined as any suitable integer or real type.

MODE NUM = INT; # type of array element #

PROC sum = ([] NUM a) NUM :
        NUM result := 0;
        FOR i FROM LWB a TO UPB a DO
            result +:= a[i]

For instance, the call

sum((3, 4, 5, 6))

returns 18.


Total of a series function. Terminate by using INT_MAX <highlightsyntax language="c"> //stdarg.h must be included for this type of function to be declared

  1. include <stdarg.h>

int total( int first, ... ) {

  int count = 0, sum = 0, i = first;
  va_list marker;
  va_start( marker, first );     /* Initialize variable arguments. */
  while( i != INT_MAX )
     sum += i;                   //increases the sum
     i = va_arg( marker, int);   //Gets the next argument
  va_end( marker );              // Resets the list
  return sum;

} </highlightsyntax>


<highlightsyntax language="cpp">

  1. include<numeric>

sum = std::accumulate(coll.begin(),coll.end(),0); </highlightsyntax>

Common Lisp

<highlightsyntax language="lisp"> (reduce #'+ list) (reduce #'+ list :initial-value 0) ; improved version that also works with empty lists </highlightsyntax>

Another way:

<highlightsyntax language="lisp"> (apply #'+ list) </highlightsyntax>


INT_MAX = 2147483647.

sum proc
	mov ecx,DWORD PTR [esp]
	add esp,4
	xor eax,eax
	add eax,DWORD PTR [esp]
	add esp,4
	cmp DWORD PTR [esp],2147483647
	jne sum_loop1
	mov DWORD PTR [esp],ecx
sum endp


<highlightsyntax language="fsharp"> Seq.sum container </highlightsyntax>


<highlightsyntax language="haskell"> sum container </highlightsyntax>


<highlightsyntax language="java122"> import java.util.*;

public class Total {

   public static int total(int... values) {
       int total = 0;
       for (int i : values) total += i;       
       return total;
   public static void main(String[] args) {
       System.out.println("Total: %d: ", total(1,2,4,40));

} </highlightsyntax>


Summing is a special case of folding the addition operator starting with zero:

# let sum = List.fold_left ( + ) 0;;
val sum : int list -> int = <fun>

For example:

# sum [1; 2; 4; 40];;
- : int = 47


<HIGHLIGHTSYNTAX language="perl"> use List::Util qw(sum); sum @arr; sum 0, @arr; # improved version that also works with empty arrays </HIGHLIGHTSYNTAX>


A function that sums the values in a sequence is built into python, its name is 'sum'. <highlightsyntax language="python"> assert sum([1,2,3]) == 6 </highlightsyntax> A parallel to the ANSI C example given above is as follows: <highlightsyntax language="python"> import itertools def total(*values):

 return sum(itertools.takewhile(sys.maxint.__cmp__, values))

assert total(1,2,3,sys.maxint) == 6 </highlightsyntax>


<highlightsyntax language="ruby"> class Array

 def sum;     inject(0) { |s, v| s += v }; end

end </highlightsyntax>


<highlightsyntax language="scheme"> (apply + list) </highlightsyntax>