Sum and product of an array: Difference between revisions
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=={{header|Perl}}== |
=={{header|Perl}}== |
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my $sum, $prod; |
my ($sum, $prod) = (0, 1); |
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my @list = (1, 2, 3); |
my @list = (1, 2, 3); |
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$sum += $_ for (@list); |
$sum += $_ for (@list); |
Revision as of 08:39, 22 January 2008
You are encouraged to solve this task according to the task description, using any language you may know.
Compute the sum and product of an array of integers.
4D
ARRAY INTEGER($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
Ada
type Int_Array is array(Integer range <>) of Integer;
array : Int_Arrayr := (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;
Define the product function
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;
This function will raise the pre-defined exception Constraint_Error if the product overflows the values represented by type Integer
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
BASIC
Interpreter: unknown
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
Compiler: 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
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; }
C++
Using the C++ standard library (STL):
#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>());
Template alternative:
// 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; }
C#
int sum = 0, prod = 1; int[] arg = { 1, 2, 3, 4, 5 }; foreach (int value in arg) { sum += value; prod *= value; }
Clean
array = {1, 2, 3, 4, 5} Sum = sum [x \\ x <-: array] Prod = foldl (*) 1 [x \\ x <-: array]
ColdFusion
<cfset myArray = listToArray("1,2,3,4,5")> #arraySum(myArray)#
Common Lisp
Interpreter: CLisp (ANSI Common Lisp)
(defparameter data #1A(1 2 3 4 5)) (values (reduce #'+ data) (reduce #'* data))
D
auto sum = 0, prod = 1; auto array = [1, 2, 3, 4, 5]; foreach(v; array) { sum += v; prod *= v; }
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;
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 }
Erlang
Using the standard libraries:
% 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).
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).
Factor
1 5 1 <range> dup sum . product . 15 120 { 1 2 3 4 } dup sum swap product 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
Groovy
[1,2,3,4,5].sum() [1,2,3,4,5].product()
[1,2,3,4,5].inject(0) { sum, val -> sum + val } [1,2,3,4,5].inject(1) { prod, val -> prod * val }
Haskell
For lists, sum and product are already defined in the Prelude:
values = [1..10] s = sum values -- the easy way p = product values s' = foldl (+) 0 values -- the hard way p' = foldl (*) 1 values
To do the same for an array, just convert it lazily to a list:
import Data.Array values = listArray (1,10) [1..10] s = sum $ elems $ values p = product $ elems $ values
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
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; } } }
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);
Lisp
Interpreter: XEmacs (beta) version 21.5.21 of February 2005 (+CVS-20050720)
(setq array [1 2 3 4 5]) (eval (concatenate 'list '(+) array)) (eval (concatenate 'list '(*) array))
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
OCaml
let a = [| 1; 2; 3; 4; 5 |] in Array.fold_left (+) 0 a Array.fold_left ( * ) 1 a
Variant, using a liked list rather than an array:
let x = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10];; List.fold_left (+) 0 x List.fold_left ( * ) 1 x
Perl
my ($sum, $prod) = (0, 1); my @list = (1, 2, 3); $sum += $_ for (@list); $prod *= $_ for (@list);
Alternate
Libraries: List::Util
use List::Util 'sum,product'; my @list = (1, 2, 3); my $s = sum @list; my $p = product @list;
Alternate
# TMTOWTDI
my $sum = 0, $prod; my @list = qw(1 2 3); map { $sum += $_ } @list; map($prod *= $_, @list);
PHP
$array = array(1,2,3,4,5,6,7,8,9); echo array_sum($array); echo array_product($array);
Pop11
Simple loop:
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;
One can alternativly use second order iterator:
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);
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.
test
:- sum([1,2,3,4,5,6,7,8,9],X). X =45; :- product([1,2,3,4,5],X). X = 120;
Python
Interpeter: Python 2.5
numbers = [1, 2, 3] total = sum(numbers) product = 1 for i in numbers: product *= i
Or functionally (faster but perhaps less clear):
from functools import reduce # Interpreter: Python 3.0 from operator import mul, add reduce(add, numbers) reduce(mul, numbers)
Raven
0 [ 1 2 3 ] each + 1 [ 1 2 3 ] each *
Ruby
arr = [1,2,3,4,5] # or ary = *1..5 sum = arr.inject { |sum, item| sum + item } # => 15 product = ary.inject{ |prod, element| prod * element } # => 120
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}
It may also be done in a classic imperative way :
var sum = 0; val product = 1 for (val x <- a) sum = sum + x for (val x <- a) product = product * x
Scheme
(define x '(1 2 3 4 5)) (apply + x) (apply * x)
A recursive solution, without the n-ary operator "trick":
(define (reduce f u L) (if (null? L) u (f (car L) (reduce f u (cdr L))))) (reduce + 0 '(1 2 3 4 5)) ;; 0 is unit for + (reduce * 1 '(1 2 3 4 5)) ;; 1 is unit for *
TODO: tail-recursive solution, like a good little Schemer)
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;
Call these functions with:
writeln(sumArray([](1, 2, 3, 4, 5))); writeln(prodArray([](1, 2, 3, 4, 5)));
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]
Standard ML
val array = [1,2,3,4,5]; foldl op+ 0 array; foldl (op*) 1 array;
Tcl
set arr [list 3 6 8] set sum [expr [join $arr +]] set prod [expr [join $arr *]]
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 .
UNIX Shell
Interpreter: NetBSD 3.0's ash
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
Interpreter: 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 $PROD
XSLT
XSLT (or XPath rather) has a few built-in functions for reducing from a collection, but product is not among them. Because of referential transparency, one must resort to recursive solutions for general iterative operations upon collections. The following code represents the array by numeric values in <price> elements in the source document.
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="text" /> <xsl:template name="sum-prod"> <xsl:param name="values" /> <xsl:param name="sum" select="0" /> <xsl:param name="prod" select="1" /> <xsl:choose> <xsl:when test="not($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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