Sum and product of an array

Compute the sum and product of an array of integers.

ACL2

<lang Lisp>(defun sum (xs)

  (if (endp xs)
      0
      (+ (first xs)
         (sum (rest xs)))))

(defun prod (xs)

  (if (endp xs)
      1
      (* (first xs)
         (prod (rest xs)))))</lang>

ActionScript

<lang actionscript>package { import flash.display.Sprite;

public class SumAndProduct extends Sprite { public function SumAndProduct() { var arr:Array = [1, 2, 3, 4, 5]; var sum:int = 0; var prod:int = 1;

for (var i:int = 0; i < arr.length; i++) { sum += arr[i]; prod *= arr[i]; }

trace("Sum: " + sum); // 15 trace("Product: " + prod); // 120 } } }</lang>

Ada

<lang ada>type Int_Array is array(Integer range <>) of Integer;

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

  Sum := Sum + array(I);

end loop;</lang> Define the product function <lang ada>function Product(Item : Int_Array) return Integer is

 Prod : Integer := 1;

begin

 for I in Item'range loop
    Prod := Prod * Item(I);
 end loop;
 return Prod;

end Product;</lang> This function will raise the predefined exception Constraint_Error if the product overflows the values represented by type Integer

ALGOL 68

<lang algol68>main:(

 INT default upb := 3;
 MODE INTARRAY = [default upb]INT;

 INTARRAY array = (1,2,3,4,5,6,7,8,9,10);
 INT sum := 0;
 FOR i FROM LWB array TO UPB array DO
    sum +:= array[i]
 OD;

 # Define the product function #
 PROC int product = (INTARRAY item)INT:
 (
   INT prod :=1;
   FOR i FROM LWB item TO UPB item DO
      prod *:= item[i]
   OD;
   prod
 ) # int product # ;
 printf(($" Sum: "g(0)$,sum,$", Product:"g(0)";"l$,int product(array)))

)</lang>

Output:

 Sum: 55, Product:3628800;

Aime

<lang aime>void compute(integer &s, integer &p, list l) {

   integer i;

   s = 0;
   p = 1;
   i = l_length(l);
   while (i) {
       i -= 1;
       s += l_q_integer(l, i);
       p *= l_q_integer(l, i);
   }

}

integer main(void) {

   integer sum, product;

   compute(sum, product, l_effect(2, 3, 5, 7, 11, 13, 17, 19));

   o_integer(sum);
   o_newline();
   o_integer(product);
   o_newline();

   return 0;

}</lang>

Output:

77
9699690

AutoHotkey

<lang AutoHotkey>numbers = 1,2,3,4,5 product := 1 loop, parse, numbers, `, { sum += A_LoopField product *= A_LoopField } msgbox, sum = %sum%`nproduct = %product%</lang>

For array input, it is easiest to "deserialize" it from a string with the split() function. <lang awk>$ awk 'func sum(s){split(s,a);r=0;for(i in a)r+=a[i];return r}{print sum($0)}' 1 2 3 4 5 6 7 8 9 10 55

$ awk 'func prod(s){split(s,a);r=1;for(i in a)r*=a[i];return r}{print prod($0)}' 1 2 3 4 5 6 7 8 9 10 3628800</lang>

Babel

<lang babel>main: { [2 3 5 7 11 13] sp }

sum! : { <- 0 -> { + } eachar } product!: { <- 1 -> { * } eachar }

sp!:

   { dup 
   sum %d cr <<
   product %d cr << }

Result: 41 30030</lang>

Perhaps better Babel:

<lang babel>main:

   { [2 3 5 7 11 13] 
   ar2ls dup cp
   <- sum_stack ->
   prod_stack
   %d cr << 
   %d cr << }

sum_stack:

   { { give  
       { + }
       { depth 1 > }
   do_while } nest }

prod_stack:

   { { give  
       { * }
       { depth 1 > }
   do_while } nest }</lang>

The nest operator creates a kind of argument-passing context - it saves whatever is on Top-of-Stack (TOS), saves the old stack, clears the stack and places the saved TOS on the new, cleared stack. This permits a section to monopolize the stack. At the end of the nest context, whatever is on TOS will be "passed back" to the original stack which will be restored.

The depth operator returns the current depth of the stack.

BASIC

Works with: FreeBASIC

<lang freebasic>dim array(5) as integer = { 1, 2, 3, 4, 5 }

dim sum as integer = 0 dim prod as integer = 1 for index as integer = lbound(array) to ubound(array)

 sum += array(index)
 prod *= array(index)

next</lang>

Applesoft BASIC

<lang ApplesoftBasic> 10 N = 5

20 S = 0:P = 1: DATA 1,2,3,4,5
30 N = N - 1: DIM A(N)
40  FOR I = 0 TO N
50  READ A(I): NEXT
60  FOR I = 0 TO N
70 S = S + A(I):P = P * A(I)
80  NEXT
90  PRINT "SUM="S,"PRODUCT="P</lang>

BBC BASIC

<lang bbcbasic> DIM array%(5)

     array%() = 1, 2, 3, 4, 5, 6
     
     PRINT "Sum of array elements = " ; SUM(array%())
     
     product% = 1
     FOR I% = 0 TO DIM(array%(),1)
       product% *= array%(I%)
     NEXT
     PRINT "Product of array elements = " ; product%</lang>

bc

<lang bc>a[0] = 3.0 a[1] = 1 a[2] = 4.0 a[3] = 1.0 a[4] = 5 a[5] = 9.00 n = 6 p = 1 for (i = 0; i < n; i++) {

   s += a[i]
   p *= a[i]

} "Sum: "; s "Product: "; p</lang>

Befunge

Works with: befungee

The program first reads the number of elements in the array, then the elements themselves (each number on a separate line) and calculates their sum. <lang Befunge>0 &>: #v_ $. @

      >1- \ & + \v
  ^              <</lang>

Bracmat

<lang bracmat>( ( sumprod

 =   sum prod num
   .   0:?sum
     & 1:?prod
     & (   !arg
         :   ?
             ( #%?num ?
             & !num+!sum:?sum
             & !num*!prod:?prod
             & ~
             )
       | (!sum.!prod)
       )
 )

& out$sumprod$(2 3 5 7 11 13 17 19) );</lang>

Output:

77.9699690

C

<lang c>/* using pointer arithmetic (because we can, I guess) */ int arg[] = { 1,2,3,4,5 }; int arg_length = sizeof(arg)/sizeof(arg[0]); int *end = arg+arg_length; int sum = 0, prod = 1; int *p;

for (p = arg; p!=end; ++p) {

  sum += *p;
  prod *= *p;

}</lang>

C++

Library: STL

<lang cpp>#include <numeric>

include <functional>

int arg[] = { 1, 2, 3, 4, 5 }; int sum = std::accumulate(arg, arg+5, 0, std::plus<int>()); // or just // std::accumulate(arg, arg + 5, 0); // since plus() is the default functor for accumulate int prod = std::accumulate(arg, arg+5, 1, std::multiplies<int>());</lang> Template alternative: <lang cpp>// this would be more elegant using STL collections 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;

} template <typename T> T prod (const T *array, const unsigned n) {

   T accum = 1;
   for (unsigned i=0; i<n; i++)
       accum *= array[i];
   return accum;

}

include <iostream>

using std::cout; using std::endl;

int main () {

   int aint[] = {1, 2, 3};
   cout << sum(aint,3) << " " << prod(aint, 3) << endl;
   float aflo[] = {1.1, 2.02, 3.003, 4.0004};
   cout << sum(aflo,4) << " " << prod(aflo,4) << endl;
   return 0;

}</lang>

C#

<lang csharp>int sum = 0, prod = 1; int[] arg = { 1, 2, 3, 4, 5 }; foreach (int value in arg) {

 sum += value;
 prod *= value;

}</lang>

Alternative using Linq (C# 3)

Works with: C# version 3

<lang csharp>int[] arg = { 1, 2, 3, 4, 5 }; int sum = arg.Sum(); int prod = arg.Aggregate((runningProduct, nextFactor) => runningProduct * nextFactor);</lang>

Chef

<lang chef>Sum and Product of Numbers as a Piece of Cake.

This recipe sums N given numbers.

Ingredients. 1 N 0 sum 1 product 1 number

Method. Put sum into 1st mixing bowl. Put product into 2nd mixing bowl. Take N from refrigerator. Chop N. Take number from refrigerator. Add number into 1st mixing bowl. Combine number into 2nd mixing bowl. Chop N until choped. Pour contents of 2nd mixing bowl into the baking dish. Pour contents of 1st mixing bowl into the baking dish.

Serves 1.</lang>

Clean

<lang clean>array = {1, 2, 3, 4, 5} Sum = sum [x \\ x <-: array] Prod = foldl (*) 1 [x \\ x <-: array]</lang>

Clojure

<lang lisp>(defn sum [vals] (reduce + vals))

(defn product [vals] (reduce * vals))</lang>

COBOL

<lang cobol> IDENTIFICATION DIVISION.

      PROGRAM-ID. array-sum-and-product.

      DATA DIVISION.
      WORKING-STORAGE SECTION.
      78  Array-Size              VALUE 10.
      01  array-area              VALUE "01020304050607080910".
          03  array               PIC 99 OCCURS Array-Size TIMES.

      01  array-sum               PIC 9(8).
      01  array-product           PIC 9(10) VALUE 1.

      01  i                       PIC 99.

      PROCEDURE DIVISION.
          PERFORM VARYING i FROM 1 BY 1 UNTIL Array-Size < i
              ADD array (i) TO array-sum
              MULTIPLY array (i) BY array-product
          END-PERFORM

          DISPLAY "Sum:     " array-sum
          DISPLAY "Product: " array-product

          GOBACK
          .</lang>

CoffeeScript

<lang coffeescript> sum = (array) ->

 array.reduce (x, y) -> x + y

product = (array) ->

 array.reduce (x, y) -> x * y

</lang>

ColdFusion

Sum of an Array, <lang cfm><cfset Variables.myArray = [1,2,3,4,5,6,7,8,9,10]> <cfoutput>#ArraySum(Variables.myArray)#</cfoutput></lang>

Product of an Array, <lang cfm><cfset Variables.myArray = [1,2,3,4,5,6,7,8,9,10]> <cfset Variables.Product = 1> <cfloop array="#Variables.myArray#" index="i">

<cfset Variables.Product *= i>

</cfloop> <cfoutput>#Variables.Product#</cfoutput></lang>

Common Lisp

<lang lisp>(let ((data #(1 2 3 4 5))) ; the array

 (values (reduce #'+ data)       ; sum
         (reduce #'* data)))     ; product</lang>

The loop macro also has support for sums. <lang lisp>(loop for i in '(1 2 3 4 5) sum i)</lang>

D

<lang d>import std.stdio;

void main() {

   immutable array = [1, 2, 3, 4, 5];

   int sum = 0;
   int prod = 1;

   foreach (x; array) {
       sum += x;
       prod *= x;
   }

   writeln("Sum: ", sum);
   writeln("Product: ", prod);

}</lang>

Output:

Sum: 15
Product: 120

Compute sum and product of array in one pass (same output): <lang d>import std.stdio, std.algorithm, std.typecons;

void main() {

   immutable array = [1, 2, 3, 4, 5];

   // Results are stored in a 2-tuple
   immutable r = reduce!(q{a + b}, q{a * b})(tuple(0, 1), array);

   writeln("Sum: ", r[0]);
   writeln("Product: ", r[1]);

}</lang>

Delphi

<lang delphi>program SumAndProductOfArray;

{$APPTYPE CONSOLE}

var

 i: integer;
 lIntArray: array [1 .. 5] of integer = (1, 2, 3, 4, 5);
 lSum: integer = 0;
 lProduct: integer = 1;

begin

 for i := 1 to length(lIntArray) do
 begin
   Inc(lSum, lIntArray[i]);
   lProduct := lProduct * lIntArray[i]
 end;

 Write('Sum: ');
 Writeln(lSum);
 Write('Product: ');
 Writeln(lProduct);

end.</lang>

E

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

Emacs Lisp

Works with: XEmacs version version 21.5.21

<lang lisp>(setq array [1 2 3 4 5]) (eval (concatenate 'list '(+) array)) (eval (concatenate 'list '(*) array))</lang>

With a list

<lang lisp>(setq array '(1 2 3 4 5)) (apply '+ array) (apply '* array)</lang>

With explicit conversion

<lang lisp>(setq array [1 2 3 4 5]) (apply '+ (append array nil)) (apply '* (append array nil))</lang>

Eiffel

<lang eiffel> note description : "project application root class" date : "$Date$" revision : "$Revision$"

class APPLICATION

inherit ARGUMENTS

create make

feature {NONE}

array : ARRAY[INTEGER]

make do create array.make_filled (0, 0, 4) array.put (2, 0) array.put (4, 1) array.put (6, 2) array.put (8, 3) array.put (10, 4)

print("%NSum of the elements of the array: ") print(sum(array)) print("%NProduct of the elements of the array: ") print(product(array)) end

sum(ar : ARRAY[INTEGER]):INTEGER local s, i: INTEGER do from i := 0 until i > 4 loop s := s + ar[i] i := i + 1 end Result := s end

product(ar : ARRAY [INTEGER]):INTEGER local prod, i: INTEGER do prod := 1 from i := 0 until i > 4 loop prod := prod * ar[i] i := i + 1 end Result := prod end end </lang>

Output:

Sum of the elements of the array: 30
Product of the elements of the array: 3840

Erlang

Using the standard libraries: <lang erlang>% create the list: L = lists:seq(1, 10).

% and compute its sum: S = lists:sum(L). P = lists:foldl(fun (X, P) -> X * P end, 1, L).</lang> To compute sum and products in one pass: <lang erlang> {Prod,Sum} = lists:foldl(fun (X, {P,S}) -> {P*X,S+X} end, {1,0}, lists:seq(1,10)).</lang> Or defining our own versions: <lang erlang>-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).</lang>

Euphoria

<lang euphoria>sequence array integer sum,prod

array = { 1, 2, 3, 4, 5 }

sum = 0 prod = 1 for i = 1 to length(array) do

 sum += array[i]
 prod *= array[i]

end for

printf(1,"sum is %d\n",sum) printf(1,"prod is %d\n",prod)</lang>

Output:

 sum is 15
 prod is 120

F#

<lang fsharp> let numbers = [| 1..10 |] let sum = numbers |> Array.sum let product = numbers |> Array.fold (*) 1 </lang>

Factor

<lang factor>1 5 1 <range> [ sum . ] [ product . ] bi

   15 120

{ 1 2 3 4 } [ sum ] [ product ] bi

   10 24</lang>

sum and product are defined in the sequences vocabulary: <lang factor>: sum ( seq -- n ) 0 [ + ] reduce ;

product ( seq -- n ) 1 [ * ] reduce ;</lang>

FALSE

Strictly speaking, there are no arrays in FALSE. However, a number of elements on the stack could be considered an array. The implementation below assumes the length of the array on top of the stack, and the actual items below it. Note that this implementation does remove the "array" from the stack, so in case the original values need to be retained, a copy should be provided before executing this logic. <lang false>1 2 3 4 5 {input "array"} 5 {length of input} 0s: {sum} 1p: {product}

[$0=~][1-\$s;+s:p;*p:]#%

"Sum: "s;." Product: "p;.</lang>

Output:

Sum: 15
Product: 120

Fantom

<lang fantom> class Main {

 public static Void main ()
 {
   Int[] array := (1..20).toList
   
   // you can use a loop
   Int sum := 0
   array.each |Int n| { sum += n }
   echo ("Sum of array is : $sum")

   Int product := 1
   array.each |Int n| { product *= n }
   echo ("Product of array is : $product")

   // or use 'reduce'
   // 'reduce' takes a function, 
   //       the first argument is the accumulated value
   //       and the second is the next item in the list
   sum = array.reduce(0) |Obj r, Int v -> Obj| 
   { 
     return (Int)r + v 
   }
   echo ("Sum of array : $sum")

   product = array.reduce(1) |Obj r, Int v -> Obj| 
   { 
     return (Int)r * v 
   }
   echo ("Product of array : $product")
 }

} </lang>

Forth

<lang forth>: third ( a b c -- a b c a ) 2 pick ;

reduce ( xt n addr cnt -- n' ) \ where xt ( a b -- n )

 cells bounds do i @ third execute  cell +loop nip ;

create a 1 , 2 , 3 , 4 , 5 ,

' + 0 a 5 reduce . \ 15 ' * 1 a 5 reduce . \ 120</lang>

Fortran

In ISO Fortran 90 and later, use SUM and PRODUCT intrinsics: <lang fortran>integer, dimension(10) :: a = (/ (i, i=1, 10) /) integer :: sresult, presult

sresult = sum(a); presult = product(a);</lang>

Frink

<lang frink> a = [1,2,3,5,7] sum[a] product[a] </lang>

GAP

<lang gap>v := [1 .. 8];

Sum(v);

36

Product(v);

40320

You can sum or multiply the result of a function

Sum(v, n -> n^2);

204

Product(v, n -> 1/n);

1/40320</lang>

Go

<lang go>package main

import "fmt"

func main() {

   sum, prod := 0, 1
   for _, x := range []int{1,2,5} {
       sum += x
       prod *= x
   }
   fmt.Println(sum, prod)

}</lang>

Groovy

Groovy adds a "sum()" method for collections, but not a "product()" method: <lang groovy>[1,2,3,4,5].sum()</lang> However, for general purpose "reduction" or "folding" operations, Groovy does provide an "inject()" method for collections similar to "inject" in Ruby. <lang groovy>[1,2,3,4,5].inject(0) { sum, val -> sum + val } [1,2,3,4,5].inject(1) { prod, val -> prod * val }</lang> You can also combine these operations: <lang groovy>println ([1,2,3,4,5].inject([sum: 0, product: 1]) { result, value ->

   [sum: result.sum + value, product: result.product * value]})</lang>

GW-BASIC

Works with: GW-BASIC

Works with: QBasic

<lang qbasic>10 REM Create an array with some test data in it 20 DIM A(5) 30 FOR I = 1 TO 5: READ A(I): NEXT I 40 DATA 1, 2, 3, 4, 5 50 REM Find the sum of elements in the array 60 S = 0 65 P = 1 70 FOR I = 1 TO 5 72 S = SUM + A(I) 75 P = P * A(I) 77 NEXT I 80 PRINT "The sum is "; S; 90 PRINT " and the product is "; P</lang>

Haskell

For lists, sum and product are already defined in the Prelude: <lang haskell>values = [1..10]

s = sum values -- the easy way p = product values

s' = foldl (+) 0 values -- the hard way p' = foldl (*) 1 values</lang> To do the same for an array, just convert it lazily to a list: <lang haskell>import Data.Array

values = listArray (1,10) [1..10]

s = sum . elems $ values p = product . elems $ values</lang>

HicEst

<lang hicest>array = $ ! 1, 2, ..., LEN(array)

sum = SUM(array)

product = 1 ! no built-in product function in HicEst DO i = 1, LEN(array)

 product = product * array(i)

ENDDO

WRITE(ClipBoard, Name) n, sum, product ! n=100; sum=5050; product=9.33262154E157;</lang>

IDL

<lang idl>array = [3,6,8] print,total(array) print,product(array)</lang>

Icon and Unicon

The program below prints the sum and product of the arguments to the program. <lang Icon>procedure main(arglist) every ( sum := 0 ) +:= !arglist every ( prod := 1 ) *:= !arglist write("sum := ", sum,", prod := ",prod) end</lang>

Inform 7

<lang inform7>Sum And Product is a room.

To decide which number is the sum of (N - number) and (M - number) (this is summing): decide on N + M.

To decide which number is the product of (N - number) and (M - number) (this is production): decide on N * M.

When play begins: let L be {1, 2, 3, 4, 5}; say "List: [L in brace notation], sum = [summing reduction of L], product = [production reduction of L]."; end the story.</lang>

J

<lang j>sum =: +/ product =: */</lang>

For example:

<lang j> sum 1 3 5 7 9 11 13 49

  product 1 3 5 7 9 11 13

135135

  a=: 3 10 ?@$ 100  NB. random array
  a

90 47 58 29 22 32 55 5 55 73 58 50 40 5 69 46 34 40 46 84 29 8 75 97 24 40 21 82 77 9

  NB. on a table, each row is an item to be summed:
  sum a

177 105 173 131 115 118 110 127 178 166

  product a

151380 18800 174000 14065 36432 58880 39270 16400 194810 55188

  NB. but we can tell J to sum everything within each row, instead:
  sum"1 a

466 472 462

  product"1 a

5.53041e15 9.67411e15 1.93356e15</lang>

Java

Works with: Java version 1.5+

<lang java5>public class SumProd {

public static void main(final String[] args)
{
 int sum = 0;
 int prod = 1;
 int[] arg = {1,2,3,4,5};
 for (int i : arg)
 {
  sum += i;
  prod *= i;
 }
}

}</lang>

Works with: Java version 1.8+

<lang java5>import java.util.Arrays;

public class SumProd {

public static void main(final String[] args)
{
 int[] arg = {1,2,3,4,5};
 System.out.printf("sum = %d\n", Arrays.stream(arg).sum());
 System.out.printf("sum = %d\n", Arrays.stream(arg).reduce(0, (a, b) -> a + b));
 System.out.printf("product = %d\n", Arrays.stream(arg).reduce(1, (a, b) -> a * b));
}

}</lang>

Output:

sum = 15
sum = 15
product = 120

JavaScript

<lang javascript>var array = [1, 2, 3, 4, 5],

   sum = 0,
   prod = 1,
   i;

for (i = 0; i < array.length; i += 1) {

   sum += array[i];
   prod *= array[i];

} alert(sum + ' ' + prod);</lang>

Works with: Javascript version 1.8

Where supported, the reduce method can also be used: <lang javascript>var array = [1, 2, 3, 4, 5],

   sum = array.reduce(function (a, b) {
       return a + b;
   }, 0),
   prod = array.reduce(function (a, b) {
       return a * b;
   }, 1);

alert(sum + ' ' + prod);</lang>

jq

The builtin filter, add/0, computes the sum of an array: <lang jq>[4,6,8] | add

=> 18</lang>

An efficient companion filter for computing the product of the items in an array can be defined as follows: <lang jq>def prod: reduce .[] as $i; (1; . * $i);</lang> Example:

[4,6,8] | prod
# => 192

Julia

<lang julia>julia> sum([4,6,8]) 18

julia> prod([4,6,8]) 192</lang>

Lang5

<lang lang5>4 iota 1 + dup

'+ reduce '* reduce</lang>

Lasso

<lang Lasso>local(x = array(1,2,3,4,5,6,7,8,9,10)) // sum of array elements 'Sum: ' with n in #x sum #n '\r' // product of arrray elements 'Product: ' local(product = 1) with n in #x do => { #product *= #n }

product</lang>

Output:

Sum: 55
Product: 3628800

Liberty BASIC

<lang lb>Dim array(19)

For i = 0 To 19

   array(i) = Int(Rnd(1) * 20)

Next i

'product must first equal one or you will get 0 as the product product = 1 For i = 0 To 19

   sum = (sum + array(i))
   product = (product * array(i))

next i

Print "Sum is " + str$(sum) Print "Product is " + str$(product)</lang>

Logo

<lang logo>print apply "sum arraytolist {1 2 3 4 5} print apply "product arraytolist {1 2 3 4 5}</lang>

Lua

<lang lua> function sumf(a, ...) return a and a + sumf(...) or 0 end function sumt(t) return sumf(unpack(t)) end function prodf(a, ...) return a and a * prodf(...) or 1 end function prodt(t) return prodf(unpack(t)) end

print(sumt{1, 2, 3, 4, 5}) print(prodt{1, 2, 3, 4, 5})</lang>

Lucid

prints a running sum and product of sequence 1,2,3... <lang lucid>[%sum,product%]

where
   x = 1 fby x + 1;
   sum = 0 fby sum + x;
   product = 1 fby product * x
end</lang>

Mathematica

Mathematica provides many ways of doing the sum of an array (any kind of numbers or symbols): <lang Mathematica>a = {1, 2, 3, 4, 5} Plus @@ a Apply[Plus, a] Total[a] Total@a a // Total Sum[ai, {i, 1, Length[a]}] Sum[i, {i, a}]</lang> all give 15. For product we also have a couple of choices: <lang Mathematica>a = {1, 2, 3, 4, 5} Times @@ a Apply[Times, a] Product[ai, {i, 1, Length[a]}] Product[i, {i, a}]</lang> all give 120.

MATLAB

These two function are built into MATLAB as the "sum(array)" and "prod(array)" functions.

Sample Usage: <lang MATLAB>>> array = [1 2 3;4 5 6;7 8 9]

array =

    1     2     3
    4     5     6
    7     8     9

>> sum(array,1)

ans =

   12    15    18

>> sum(array,2)

ans =

    6
   15
   24

>> prod(array,1)

ans =

   28    80   162

>> prod(array,2)

ans =

    6
  120
  504</lang>

Maxima

<lang maxima>lreduce("+", [1, 2, 3, 4, 5, 6, 7, 8]); 36

lreduce("*", [1, 2, 3, 4, 5, 6, 7, 8]); 40320</lang>

MAXScript

<lang maxscript>arr = #(1, 2, 3, 4, 5) sum = 0 for i in arr do sum += i product = 1 for i in arr do product *= i</lang>

МК-61/52

Instruction: РX - array length, Р1:РC - array, РD and РE - sum and product of an array.

Modula-3

<lang modula3>MODULE Sumprod EXPORTS Main;

FROM IO IMPORT Put; FROM Fmt IMPORT Int;

VAR a := ARRAY [1..5] OF INTEGER {1, 2, 3, 4, 5}; VAR sum: INTEGER := 0; VAR prod: INTEGER := 1;

BEGIN

 FOR i := FIRST(a) TO LAST(a) DO
   INC(sum, a[i]);
   prod := prod * a[i];
 END;
 Put("Sum of array: " & Int(sum) & "\n");
 Put("Product of array: " & Int(prod) & "\n");

END Sumprod.</lang>

Output:

Sum of array: 15
Product of array: 120

MUMPS

<lang MUMPS> SUMPROD(A)

;Compute the sum and product of the numbers in the array A
NEW SUM,PROD,POS
;SUM is the running sum, 
;PROD is the running product,
;POS is the position within the array A
SET SUM=0,PROD=1,POS=""
FOR  SET POS=$ORDER(A(POS)) Q:POS=""  SET SUM=SUM+A(POS),PROD=PROD*A(POS)
WRITE !,"The sum of the array is "_SUM
WRITE !,"The product of the array is "_PROD
KILL SUM,PROD,POS
QUIT</lang>

Example:

USER>SET C(-1)=2,C("A")=3,C(42)=1,C(0)=7
 
USER>D SUMPROD^ROSETTA(.C)
 
The sum of the array is 13
The product of the array is 42

Note - the string "A" converts to 0 when doing mathematical operations.

USER>SET C(-1)=2,C("A")="3H",C(42)=.1,C(0)=7.0,C("B")="A"
 
USER>D SUMPROD^ROSETTA(.C)
 
The sum of the array is 12.1
The product of the array is 0

Nemerle

As mentioned for some of the other functional languages, it seems more natural to work with lists in Nemerle, but as the task specifies working on an array, this solution will work on either. <lang Nemerle>using System; using System.Console; using System.Collections.Generic; using Nemerle.Collections;

module SumProd {

   Sum[T] (nums : T) : int
     where T : IEnumerable[int]
   {
       nums.FoldLeft(0, _+_)
   }
   
   Product[T] (nums : T) : int
     where T : IEnumerable[int]
   {
       nums.FoldLeft(1, _*_)
   }
   
   Main() : void
   {
       def arr = array[1, 2, 3, 4, 5];
       def lis = [1, 2, 3, 4, 5];
       
       def suml = Sum(lis);
       def proda = Product(arr);
       
       WriteLine("Sum is: {0}\tProduct is: {1}", suml, proda);
   }

}</lang>

NetRexx

<lang NetRexx>/* NetRexx */

options replace format comments java crossref savelog symbols binary

harry = [long 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

sum = long 0 product = long 1 entries = Rexx

loop n_ = int 0 to harry.length - 1

 nxt = harry[n_]
 entries = entries nxt
 sum = sum + nxt
 product = product * nxt 
 end n_

entries = entries.strip

say 'Sum and product of' entries.changestr(' ', ',')':' say ' Sum:' sum say ' Product:' product

return </lang>

Output:

 Sum and product of 1,2,3,4,5,6,7,8,9,10:
     Sum: 55
 Product: 3628800

NewLISP

<lang NewLISP>(setq a '(1 2 3 4 5)) (apply + a) (apply * a)</lang>

Nial

Nial being an array language, what applies to individual elements are extended to cover array operations by default strand notation <lang nial>+ 1 2 3 = 6

1 2 3

= 6</lang> array notation <lang nial>+ [1,2,3]</lang> grouped notation <lang nial>(* 1 2 3) = 6

(1 2 3)

= 6</lang> (All these notations are equivalent)

Nim

<lang nim>var xs = @[1,2,3,4,5,6]

var sum, product: int

product = 1

for x in xs:

 sum += x
 product *= x</lang>

Or functionally: <lang nim>import sequtils

let

 xs = @[1,2,3,4,5,6]
 sum = xs.foldl(a + b)
 product = xs.foldl(a * b)</lang>

Objeck

<lang objeck> sum := 0; prod := 1; arg := [1, 2, 3, 4, 5]; each(i : arg) {

 sum += arg[i];
 prod *= arg[i];

}; </lang>

Objective-C

Works with: GCC version 4.0.1 (apple)

Sum: <lang objc>- (float) sum:(NSMutableArray *)array { int i, sum, value; sum = 0; value = 0;

for (i = 0; i < [array count]; i++) { value = [[array objectAtIndex: i] intValue]; sum += value; }

return suml; }</lang> Product: <lang objc>- (float) prod:(NSMutableArray *)array { int i, prod, value; prod = 0; value = 0;

for (i = 0; i < [array count]; i++) { value = [[array objectAtIndex: i] intValue]; prod *= value; }

return suml; }</lang>

OCaml

Arrays

<lang ocaml>(* ints *) let a = [| 1; 2; 3; 4; 5 |];; Array.fold_left (+) 0 a;; Array.fold_left ( * ) 1 a;; (* floats *) let a = [| 1.0; 2.0; 3.0; 4.0; 5.0 |];; Array.fold_left (+.) 0.0 a;; Array.fold_left ( *.) 1.0 a;;</lang>

Lists

<lang ocaml>(* ints *) let x = [1; 2; 3; 4; 5];; List.fold_left (+) 0 x;; List.fold_left ( * ) 1 x;; (* floats *) let x = [1.0; 2.0; 3.0; 4.0; 5.0];; List.fold_left (+.) 0.0 x;; List.fold_left ( *.) 1.0 x;;</lang>

Octave

<lang octave>a = [ 1, 2, 3, 4, 5, 6 ]; b = [ 10, 20, 30, 40, 50, 60 ]; vsum = a + b; vprod = a .* b;</lang>

Oforth

<lang Oforth>[1, 2, 3, 4, 5 ] sum println [ 1, 3, 5, 7, 9 ] prod println</lang>

Output:

15
945

Oz

Calculations like this are typically done on lists, not on arrays: <lang oz>declare

 Xs = [1 2 3 4 5]
 Sum = {FoldL Xs Number.'+' 0}
 Product = {FoldL Xs Number.'*' 1}

in

 {Show Sum}
 {Show Product}</lang>

If you are actually working with arrays, a more imperative approach seems natural: <lang oz>declare

 Arr = {Array.new 1 3 0}
 Sum = {NewCell 0}

in

 Arr.1 := 1
 Arr.2 := 2
 Arr.3 := 3

 for I in {Array.low Arr}..{Array.high Arr} do
    Sum := @Sum + Arr.I
 end
 {Show @Sum}</lang>

PARI/GP

<lang parigp>vecsum(v)={

 sum(i=1,#v,v[i])

}; vecprod(v)={

 prod(i=1,#v,v[i])

};</lang>

Pascal

See Delphi

Perl

<lang perl>my @list = ( 1, 2, 3 );

my ( $sum, $prod ) = ( 0, 1 ); $sum += $_ foreach @list; $prod *= $_ foreach @list;</lang> Or using the List::Util module: <lang perl>use List::Util qw/sum0 product/; my @list = (1..9);

say "Sum: ", sum0(@list); # sum0 returns 0 for an empty list say "Product: ", product(@list);</lang>

Output:

Sum: 45
Product: 362880

Perl 6

<lang perl6>my @ary = 1, 5, 10, 100; say 'Sum: ', [+] @ary; say 'Product: ', [*] @ary;</lang>

PHP

<lang php>$array = array(1,2,3,4,5,6,7,8,9); echo array_sum($array); echo array_product($array);</lang>

PicoLisp

<lang PicoLisp>(let Data (1 2 3 4 5)

  (cons
     (apply + Data)
     (apply * Data) ) )</lang>

Output:

(15 . 120)

PL/I

<lang pli>declare A(10) fixed binary static initial

  (1, 2, 3, 4, 5, 6, 7, 8, 9, 10);

put skip list (sum(A)); put skip list (prod(A));</lang>

Pop11

Simple loop: <lang pop11>lvars i, sum = 0, prod = 1, ar = {1 2 3 4 5 6 7 8 9}; for i from 1 to length(ar) do

   ar(i) + sum -> sum;
   ar(i) * prod -> prod;

endfor;</lang> One can alternatively use second order iterator: <lang pop11>lvars sum = 0, prod = 1, ar = {1 2 3 4 5 6 7 8 9}; appdata(ar, procedure(x); x + sum -> sum; endprocedure); appdata(ar, procedure(x); x * prod -> prod; endprocedure);</lang>

PowerShell

The Measure-Object cmdlet already knows how to compute a sum: <lang powershell>function Get-Sum ($a) {

   return ($a | Measure-Object -Sum).Sum

}</lang> But not how to compute a product: <lang powershell>function Get-Product ($a) {

   if ($a.Length -eq 0) {
       return 0
   } else {
       $p = 1
       foreach ($x in $a) {
           $p *= $x
       }
       return $p
   }

}</lang> One could also let PowerShell do all the work by simply creating an expression to evaluate:

Works with: PowerShell version 2

<lang powershell>function Get-Product ($a) {

   if ($a.Length -eq 0) {
       return 0
   }
   $s = $a -join '*'
   return (Invoke-Expression $s)

}</lang> Even nicer, however, is a function which computes both at once and returns a custom object with appropriate properties: <lang powershell>function Get-SumAndProduct ($a) {

   $sum = 0
   if ($a.Length -eq 0) {
       $prod = 0
   } else {
       $prod = 1
       foreach ($x in $a) {
           $sum += $x
           $prod *= $x
       }
   }
   $ret = New-Object PSObject
   $ret | Add-Member NoteProperty Sum $sum
   $ret | Add-Member NoteProperty Product $prod
   return $ret

}</lang>

Output:

PS> Get-SumAndProduct 5,9,7,2,3,8,4

Sum Product
--- -------
 38   60480

Prolog

<lang prolog>sum([],0). sum([H|T],X) :- sum(T,Y), X is H + Y. product([],1). product([H|T],X) :- product(T,Y), X is H * X.</lang>

test

:- sum([1,2,3,4,5,6,7,8,9],X).
X =45;
:- product([1,2,3,4,5],X).
X = 120;

Using fold <lang prolog> add(A,B,R):-

   R is A + B.

mul(A,B,R):-

   R is A * B.

% define fold now. fold([], Act, Init, Init).

fold(Lst, Act, Init, Res):-

   head(Lst,Hd),
   tail(Lst,Tl),
   apply(Act,[Init, Hd, Ra]),
   fold(Tl, Act, Ra, Res).

sumproduct(Lst, Sum, Prod):-

   fold(Lst,mul,1, Prod),
   fold(Lst,add,0, Sum).

?- sumproduct([1,2,3,4],Sum,Prod). Sum = 10, Prod = 24 .

</lang>

PostScript

<lang> /sumandproduct { /x exch def /sum 0 def /prod 0 def /i 0 def x length 0 eq { } { /prod prod 1 add def x length{ /sum sum x i get add def /prod prod x i get mul def /i i 1 add def }repeat }ifelse sum == prod == }def </lang>

Library: initlib

<lang postscript> % sum [1 1 1 1 1] 0 {add} fold % product [1 1 1 1 1] 1 {mul} fold

</lang>

PureBasic

<lang PureBasic>Dim MyArray(9) Define a, sum=0, prod=1

For a = 0 To ArraySize(MyArray()) ; Create a list of some random numbers

 MyArray(a) = 1 + Random(9)          ; Insert a number [1...10] in current element

 sum  + MyArray(a)
 prod * MyArray(a)

Python

Works with: Python version 2.5

<lang python>numbers = [1, 2, 3] total = sum(numbers)

product = 1 for i in numbers:

   product *= i</lang>

Or functionally (faster but perhaps less clear):

Works with: Python version 2.5

<lang python>from operator import mul, add sum = reduce(add, numbers) # note: this version doesn't work with empty lists sum = reduce(add, numbers, 0) product = reduce(mul, numbers) # note: this version doesn't work with empty lists product = reduce(mul, numbers, 1)</lang>

Library: numpy

<lang python>from numpy import r_ numbers = r_[1:4] total = numbers.sum() product = numbers.prod()</lang>

If you are summing floats in Python 2.6+, you should use math.fsum() to avoid loss of precision:

Works with: Python version 2.6, 3.x

<lang python>import math total = math.fsum(floats)</lang>

R

<lang r>total <- sum(1:5) product <- prod(1:5)</lang>

Racket

lang racket

(for/sum ([x #(3 1 4 1 5 9)]) x) (for/product ([x #(3 1 4 1 5 9)]) x) </lang>

Raven

REBOL

<lang REBOL>REBOL [

   Title: "Sum and Product"
   Date: 2010-01-04
   Author: oofoe
   URL: http://rosettacode.org/wiki/Sum_and_product_of_array

]

Simple

sum: func [a [block!] /local x] [x: 0 forall a [x: x + a/1] x]

product: func [a [block!] /local x] [x: 1 forall a [x: x * a/1] x]

Way too fancy

redux: func [ "Applies an operation across an array to produce a reduced value." a [block!] "Array to operate on." op [word!] "Operation to perform." /init x "Initial value (default 0)." ][if not init [x: 0] forall a [x: do compose [x (op) (a/1)]] x]

rsum: func [a [block!]][redux a '+]

rproduct: func [a [block!]][redux/init a '* 1]

Tests

assert: func [code][print [either do code [" ok"]["FAIL"] mold code]]

print "Simple dedicated functions:" assert [55 = sum [1 2 3 4 5 6 7 8 9 10]] assert [3628800 = product [1 2 3 4 5 6 7 8 9 10]]

print [crlf "Fancy reducing function:"] assert [55 = rsum [1 2 3 4 5 6 7 8 9 10]] assert [3628800 = rproduct [1 2 3 4 5 6 7 8 9 10]]</lang>

Output:

Simple dedicated functions:
  ok [55 = sum [1 2 3 4 5 6 7 8 9 10]]
  ok [3628800 = product [1 2 3 4 5 6 7 8 9 10]]

Fancy reducing function:
  ok [55 = rsum [1 2 3 4 5 6 7 8 9 10]]
  ok [3628800 = rproduct [1 2 3 4 5 6 7 8 9 10]]

REXX

<lang rexx>/*REXX program to add and separately multiply elements of an array. */ numeric digits 30 /*allow 30-digit numbers (default is 9)*/ m=20 /*one method of indicating array size. */

         do j=1  for m      /*build an array of twenty elements.   */
         y.j=j              /*set 1st to 1, 3rd to 3, 9th to 9 ... */
         end   /*j*/

sum=0 /*initialize SUM to zero. */ prod=1 /*initialize PROD to unity. */

         do k=1  for m
         sum =sum +y.k      /*add the element to the running total.*/    
         prod=prod*y.k      /*multiple the element to running prod.*/
         end   /*k*/

say ' sum of' m "elements for the Y array is: " sum say 'product of' m "elements for the Y array is: " prod

                            /*stick a fork in it, we're done.      */</lang>

Output:

    sum of 20 elements for the Y array is:  210
product of 20 elements for the Y array is:  2432902008176640000

Ruby

<lang ruby>arr = [1,2,3,4,5] # or ary = *1..5, or ary = (1..5).to_a p sum = arr.inject(0) { |sum, item| sum + item }

=> 15

p product = arr.inject(1) { |prod, element| prod * element }

=> 120</lang>

Works with: Ruby version 1.8.7

<lang ruby>arr = [1,2,3,4,5] p sum = arr.inject(0, :+) #=> 15 p product = arr.inject(1, :*) #=> 120

If you do not explicitly specify an initial value for memo,
then the first element of collection is used as the initial value of memo.

p sum = arr.inject(:+) #=> 15 p product = arr.inject(:*) #=> 120</lang>

Note: When the Array is empty, the initial value returns. However, nil returns if not giving an initial value. <lang ruby>arr = [] p arr.inject(0, :+) #=> 0 p arr.inject(1, :*) #=> 1 p arr.inject(:+) #=> nil p arr.inject(:*) #=> nil</lang>

Enumerable#reduce is the alias of Enumerable#inject.

Run BASIC

<lang runbasic>dim array(100) for i = 1 To 100

   array(i) = rnd(0) * 100

next i

product = 1 for i = 0 To 19

   sum     = (sum + array(i))
   product = (product * array(i))

next i

Print " Sum is ";sum Print "Product is ";product</lang>

Rust

<lang rust>use std::iter::{AdditiveIterator, MultiplicativeIterator};

fn main() {

 let arr = [1i, 2, 3, 4, 5, 6, 7, 8, 9];
 
 // using fold
 let sum = arr.iter().fold(0, |a, &b| a + b);
 let product = arr.iter().fold(1, |a, &b| a * b);
 println!("the sum is {:d} and the product is {:d}", sum, product);
 
 // or using sum and product from AdditiveIterator
 // and MultiplicativeIterator
 let sum2 = arr.iter().map(|&a| a).sum();
 let product2 = arr.iter().map(|&a| a).product();
 println!("the sum is {:d} and the product is {:d}", sum2, product2);

}</lang>

SAS

<lang sas>data _null_;

  array a{*} a1-a100;
  do i=1 to 100;
     a{i}=i*i;
  end;
  b=sum(of a{*});
  put b c;

run;</lang>

Sather

<lang sather>class MAIN is

 main is
   a :ARRAY{INT} := |10, 5, 5, 20, 60, 100|;
   sum, prod :INT;
   loop sum := sum + a.elt!; end;
   prod := 1;
   loop prod := prod * a.elt!; end;
   #OUT + sum + " " + prod + "\n";
 end;

end;</lang>

Scala

<lang scala>val seq = Seq(1, 2, 3, 4, 5) val sum = seq.foldLeft(0)(_ + _) val product = seq.foldLeft(1)(_ * _)</lang>

Or even shorter: <lang scala>val sum = seq.sum val product = seq.product</lang>

Works with all data types for which a Numeric implicit is available.

Scheme

<lang scheme>(apply + '(1 2 3 4 5)) (apply * '(1 2 3 4 5))</lang> A tail-recursive solution, without the n-ary operator "trick". Because Scheme supports tail call optimization, this is as space-efficient as an imperative loop. <lang scheme>(define (reduce f i l)

 (if (null? l)
   i
   (reduce f (f i (car l)) (cdr l))))

(reduce + 0 '(1 2 3 4 5)) ;; 0 is unit for + (reduce * 1 '(1 2 3 4 5)) ;; 1 is unit for *</lang>

Seed7

<lang 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;

const func integer: prodArray (in array integer: valueArray) is func

 result
   var integer: prod is 1;
 local
   var integer: value is 0;
 begin
   for value range valueArray do
     prod *:= value;
   end for;
 end func;</lang>

Call these functions with:

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

SETL

<lang SETL>numbers := [1 2 3 4 5 6 7 8 9]; print(+/ numbers, */ numbers);</lang>

=> 45 362880

Sidef

<lang ruby>var ary = [1, 5, 10, 100]; say ('Sum: ', ary«+»); say ('Product: ', ary«*»);</lang>

Slate

<lang slate>#(1 2 3 4 5) reduce: [:sum :number | sum + number]

(1 2 3 4 5) reduce: [:product :number | product * number]</lang>

Shorthand for the above with a macro: <lang slate>#(1 2 3 4 5) reduce: #+ `er

(1 2 3 4 5) reduce: #* `er</lang>

Smalltalk

<lang smalltalk>#(1 2 3 4 5) inject: 0 into: [:sum :number | sum + number]

(1 2 3 4 5) inject: 1 into: [:product :number | product * number]</lang>

Some implementation also provide a fold: message: <lang smalltalk>#(1 2 3 4 5) fold: [:sum :number | sum + number]

(1 2 3 4 5) fold: [:product :number | product * number]</lang>

SNOBOL4

<lang snobol> t = table()

read the integer from the std. input

init_tab t<x = x + 1> = trim(input) :s(init_tab)

         product = 1
         sum = 0

counting backwards to 1

loop i = t< x = ?gt(x,1) x - 1> :f(out)

         sum = sum + i
         product = product * i         :(loop)

out output = "Sum: " sum

         output = "Prod: " product

end</lang>

Input

Output:

 Sum:  15
 Prod: 120

Standard ML

Arrays

<lang sml>(* ints *) val a = Array.fromList [1, 2, 3, 4, 5]; Array.foldl op+ 0 a; Array.foldl op* 1 a; (* reals *) val a = Array.fromList [1.0, 2.0, 3.0, 4.0, 5.0]; Array.foldl op+ 0.0 a; Array.foldl op* 1.0 a;</lang>

Lists

<lang sml>(* ints *) val x = [1, 2, 3, 4, 5]; foldl op+ 0 x; foldl op* 1 x; (* reals *) val x = [1.0, 2.0, 3.0, 4.0, 5.0]; foldl op+ 0.0 x; foldl op* 1.0 x;</lang>

Sparkling

<lang Sparkling>spn:1> reduce({ 1, 2, 3, 4, 5 }, 0, function(x, y) { return x + y; }) = 15 spn:2> reduce({ 1, 2, 3, 4, 5 }, 1, function(x, y) { return x * y; }) = 120</lang>

Swift

<lang swift>let a = [1, 2, 3, 4, 5] println(a.reduce(0, +)) // prints 15 println(a.reduce(1, *)) // prints 120

println(reduce(a, 0, +)) // prints 15 println(reduce(a, 1, *)) // prints 120</lang>

Tcl

Works with: Tcl version 8.5

<lang tcl>set arr [list 3 6 8] set sum [tcl::mathop::+ {*}$arr] set prod [tcl::mathop::* {*}$arr]</lang>

TI-83 BASIC

Use the built-in functions

sum()

and

prod()

.

Toka

<lang toka>4 cells is-array foo

212 1 foo array.put 51 2 foo array.put 12 3 foo array.put 91 4 foo array.put

[ ( array size -- sum )

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

( product )

reset 1 4 0 [ i foo array.get * ] countedLoop .</lang>

Trith

<lang trith>[1 2 3 4 5] 0 [+] foldl</lang> <lang trith>[1 2 3 4 5] 1 [*] foldl</lang>

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT list="1'2'3'4'5" sum=SUM(list) PRINT " sum: ",sum

product=1 LOOP l=list product=product*l ENDLOOP PRINT "product: ",product </lang>

Output:

    sum: 15
product: 120

UNIX Shell

Works with: NetBSD version 3.0

From an internal variable, $IFS delimited:

From the argument list (ARGV):

From STDIN, one integer per line:

<lang bash>sum=0 prod=1 while read n do sum="$(($sum + $n))"; prod="$(($prod * $n))" done echo $sum $prod</lang>

Works with: GNU bash version 3.2.0(1)-release (i386-unknown-freebsd6.1)

From variable:

<lang bash>LIST='20 20 2'; SUM=0; PROD=1; for i in $LIST; do

 SUM=$[$SUM + $i]; PROD=$[$PROD * $i];

done; echo $SUM $PROD</lang>

UnixPipes

Uses ksh93-style process substitution.

Works with: bash

<lang bash>prod() {

  (read B; res=$1; test -n "$B" && expr $res \* $B || echo $res)

}

sum() {

  (read B; res=$1; test -n "$B" && expr $res + $B || echo $res)

}

fold() {

  (func=$1; while read a ; do fold $func | $func $a ; done)

}

(echo 3; echo 1; echo 4;echo 1;echo 5;echo 9) |

 tee >(fold sum) >(fold prod) > /dev/null</lang>

There is a race between fold sum and fold prod, which run in parallel. The program might print sum before product, or print product before sum.

Ursala

The reduction operator, :-, takes an associative binary function and a constant for the empty case. Natural numbers are unsigned and of unlimited size. <lang Ursala>#import nat

cast %nW

sp = ^(sum:-0,product:-1) <62,43,46,40,29,55,51,82,59,92,48,73,93,35,42,25></lang>

Output:

(875,2126997171723931187788800000)

V

<lang v>[sp dup 0 [+] fold 'product=' put puts 1 [*] fold 'sum=' put puts].</lang>

Using it:

<lang v>[1 2 3 4 5] sp = product=15 sum=120</lang>

Vala

<lang vala> public static void main(){ int sum = 0, product = 1;

int[] array = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

foreach (int number in array){ sum += number; product *= number; } } </lang>

Wart

<lang wart>def (sum_prod nums)

 (list (+ @nums) (* @nums))</lang>

XPL0

<lang XPL0>code CrLf=9, IntOut=11;

func SumProd(A, L); int A, L; int S, P, I; [S:= 0; P:= 1; for I:= 0 to L-1 do [S:= S+A(I); P:= P*A(I)]; IntOut(0, S); CrLf(0); IntOut(0, P); CrLf(0); ]; \SumSq

SumProd([1,2,3,4,5,6,7,8,9,10], 10)</lang>

Output:

55
3628800

Wortel

<lang wortel>@sum [1 2 3 4] ; returns 10 @prod [1 2 3 4] ; returns 24</lang>

XSLT

XSLT (or XPath rather) has a few built-in functions for reducing from a collection, but product is not among them. Because of referential transparency, one must resort to recursive solutions for general iterative operations upon collections. The following code represents the array by numeric values in <price> elements in the source document.

 <xsl:output method="text" />
 
 <xsl:template name="sum-prod">
   <xsl:param name="values" />
   <xsl:param name="sum"  select="0" />
   <xsl:param name="prod" select="1" />
   <xsl:choose>
     <xsl:when test="not($values)">
       <xsl:text>

Sum: </xsl:text>

       <xsl:value-of select="$sum" />
       <xsl:text>

Product: </xsl:text>

       <xsl:value-of select="$prod" />
     </xsl:when>
     <xsl:otherwise>
       <xsl:call-template name="sum-prod">
         <xsl:with-param name="values" select="$values[position() > 1]" />
         <xsl:with-param name="sum"  select="$sum  + $values[1]" />
         <xsl:with-param name="prod" select="$prod * $values[1]" />
       </xsl:call-template>
     </xsl:otherwise>
   </xsl:choose>
 </xsl:template>
 
 <xsl:template match="/">
    <xsl:text>

Sum (built-in): </xsl:text>

   <xsl:value-of select="sum(//price)" />
   
   <xsl:call-template name="sum-prod">
     <xsl:with-param name="values" select="//price" />
   </xsl:call-template>
 </xsl:template>

</xsl:stylesheet></lang>

zkl

Translation of: Clojure

<lang zkl>fcn sum(vals){vals.reduce('+,0)} fcn product(vals){vals.reduce('*,1)}</lang>

sum(T(1,2,3,4))     //-->10
product(T(1,2,3,4)) //-->24