Sum and product of an array

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Revision as of 10:17, 17 September 2009 by rosettacode>Dkf (→‎Lists: Typofix)
Task
Sum and product of an array
You are encouraged to solve this task according to the task description, using any language you may know.

Compute the sum and product of an array of integers.

4D

<lang 4d>ARRAY INTEGER($list;0) For ($i;1;5) 

      APPEND TO ARRAY($list;$i)

End for

$sum:=0 $product:=1 For ($i;1;Size of array($list)) 

  $sum:=$var+$list{$i}
  $product:=$product*$list{$i}

End for</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;

APL

Works with: APL2
      sum  ←  +/
      prod ←  ×/
      
      list ←  1 2 3 4 5 
      
      sum  list
15
      
      prod list
120

AppleScript

<lang applescript>set array to {1, 2, 3, 4, 5} set sum to 0 set product to 1 repeat with i in array 

   set sum to sum + i
   set product to product * i

end repeat</lang>

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>

AWK

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>

BASIC

Works with: QuickBasic version 4.5

<lang qbasic>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 65 PRODUCT = 1 70 FOR I = 1 TO 5 72 SUM = SUM + ARRAY(I) 75 PRODUCT = PRODUCT + ARRAY(I) 77 NEXT I 80 PRINT "The sum is ";SUM; 90 PRINT " and the product is ";PRODUCT</lang>

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>

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>

  1. 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;

}

  1. 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>

Clean

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

ColdFusion

Sum of an Array, 

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

Product of an Array, 

<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>

Common Lisp

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

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

D

<lang d>auto sum = 0, prod = 1; auto array = [1, 2, 3, 4, 5]; foreach(v; array) { 

   sum += v;
   prod *= v;

}</lang> Compute sum and product of array in one pass using std.algorithm: <lang d>auto array = [1, 2, 3, 4, 5]; auto r = reduce!("a + b", "a * b")(0, 1, array); // 0 and 1 are seeds for corresponding functions writefln("Sum: ", r._0); // Results are stored in a tuple writefln("Product: ", r._1); </lang>

Delphi

<lang delphi>var 

 Ints   : array[1..5] of integer = (1,2,3,4,5) ;
 i,Sum  : integer = 0 ;
 Prod   : integer = 1 ;

begin 

 for i := 1 to length(ints) do begin
   inc(sum,ints[i]) ;
   prod := prod * ints[i]
 end;

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
 (setq array [1 2 3 4 5])
 (eval (concatenate 'list '(+) array))
 (eval (concatenate 'list '(*) array))

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> 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>

Factor

1 5 1 <range> [ sum . ] [ product . ] bi
    15 120
{ 1 2 3 4 } [ sum ] [ product ] bi
    10 24

sum and product are defined in the sequences vocabulary: 

   : sum ( seq -- n ) 0 [+] reduce ;
   : product ( seq -- n ) 1 [ * ] reduce ;

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

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>

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>

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>

IDL

 array = [3,6,8]
 print,total(array)
 print,product(array)

J

sum=: +/
product=: */

For example: 

   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

   sum a
177 105 173 131 115 118 110 127 178 166
   product a
151380 18800 174000 14065 36432 58880 39270 16400 194810 55188

   sum"1 a
466 472 462
   product"1 a
5.53041e15 9.67411e15 1.93356e15

Java

Works with: Java version 1.5+

<lang java5>public class SumProd{ 

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

}</lang>

JavaScript

<lang javascript>var array = [1, 2, 3, 4, 5]; var sum = 0, prod = 1; for(var i in array) { 

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

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

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

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>

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)

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

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; 2; 3; 4; 5];; 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>

Oz

<lang oz>functor import 

 Application System

define 

 Print = System.showInfo

 Arr = [1 2 3 4 5]

 {Print {FoldL Arr Number.'+' 0}}
 {Print {FoldL Arr Number.'*' 1}}

 {Application.exit 0}

end</lang>

Perl

<lang perl>my ($sum, $prod) = (0, 1); my @list = (1, 2, 3); $sum += $_ foreach @list; $prod *= $_ foreach @list;</lang> Alternate:

Library: List::UtilUtil

<lang perl>use List::Util qw(sum reduce);

my @list = (1, 2, 3); my $sum1 = sum 0, @list; # 0 identity to allow empty list my $sum2 = reduce { $a + $b } 0, @list; my $product = reduce { $a * $b } 1, @list;</lang> Alternate # TMTOWTDI <lang perl>my ($sum, $prod) = (0, 1); my @list = qw(1 2 3); map { $sum += $_ } @list; map($prod *= $_, @list);</lang>

PHP

<lang php>$array = array(1,2,3,4,5,6,7,8,9); echo array_sum($array); echo array_product($array);</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 alternativly use second order iterator: <lanf 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;

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>

R

<lang r>arr <- c(1,2,3,4,5) total <- sum(arr) product <- prod(arr)</lang>

Raven

<lang raven>0 [ 1 2 3 ] each + 1 [ 1 2 3 ] each *</lang>

Ruby

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

  1. => 15

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

  1. => 120</lang>
Works with: Ruby version 1.9

<lang ruby>arr = [1,2,3,4,5] sum = arr.inject(0, :+)

  1. => 15

product = arr.inject(1, :*)

  1. => 120</lang>

Scala

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

   // (_ * _) is a shortcut for  {(x,y) => x * y}</lang>

It may also be done in a classic imperative way : <lang scala>var sum = 0; val product = 1 for (val x <- a) sum = sum + x for (val x <- a) product = product * x</lang>

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

Slate

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

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

Smalltalk

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

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

Standard ML

Arrays

<lang ocaml>(* 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 ocaml>(* 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>

Tcl

<lang tcl>set arr [list 3 6 8] set sum [expr [join $arr +]] set prod [expr [join $arr *]]</lang>

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>

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>

UNIX Shell

Works with: NetBSD version 3.0

From an internal variable, $IFS delimited: 

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

From the argument list (ARGV): 

 sum=0
 prod=1
 for n
 do sum="$(($sum + $n))"; prod="$(($prod * $n))"
 done
 echo $sum $prod

From STDIN, one integer per line: 

 sum=0
 prod=1
 while read n
 do sum="$(($sum + $n))"; prod="$(($prod * $n))"
 done
 echo $sum $prod
Works with: GNU bash version 3.2.0(1)-release (i386-unknown-freebsd6.1)

From variable: 

 LIST='20 20 2';
 SUM=0; PROD=1;
 for i in $LIST; do
   SUM=$[$SUM + $i]; PROD=$[$PROD * $i];
 done;
 echo $SUM $PROD

UnixPipes

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

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

  1. 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

[sp dup 0 [+] fold 'product=' put puts 1 [*] fold 'sum=' put puts].

Using it 

[1 2 3 4 5] sp
=
product=15
sum=120

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:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <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>
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