home > 写経編 > 柴田望洋『明解C言語 入門編』 > 6. 関数 >

ForNext

Only Do What Only You Can Do

051. 多次元配列の受け渡し

VBScript

JScript

Perl

更新日 : 2010.10.18
sub mat_add
{
    ($ma, $mb, $mc) = @_;
    
    foreach $i(0..1)
    {
        foreach $j(0..2)
        {
            $$mc[$i][$j] = $ma[$i][$j] + $mb[$i][$j];
        }
    }
}

@ma  = ([1, 2, 3], [4, 5, 6]);
@mb  = ([6, 3, 4], [5, 1, 2]);
@mc  = ([], []);

&mat_add(\@ma, \@mb, \@mc);

foreach $i(0..1)
{
    foreach $j(0..2)
    {
        printf("%3d", $mc[$i][$j]);
    }
    print "\n";
}
L:\>perl lesson_06_051.pl
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PHP

更新日 : 2010.11.03
<?php
function mat_add($ma, $mb, &$mc)
{
    for ($i = 0; $i < 2; $i++)
    {
        for ($j = 0; $j < 3; $j++)
        {
            $mc[$i][$j] = $ma[$i][$j] + $mb[$i][$j];
        }
    }
}

$ma  = array(array(1, 2, 3), array(4, 5, 6));
$mb  = array(array(6, 3, 4), array(5, 1, 2));
$mc  = array(array(),        array());

mat_add($ma, $mb, $mc);

for ($i = 0; $i < 2; $i++)
{
    for ($j = 0; $j < 3; $j++)
    {
        printf("%3d", $mc[$i][$j]);
    }
    print "\n";
}
?>
L:\>php lesson_06_051.php
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Python

更新日 : 2010.11.17
def mat_add(ma, mb, mc):
    for i in range(0, 2, 1):
        for j in range(0, 3, 1):
            mc[i][j] = ma[i][j] + mb[i][j]

ma  = [[1, 2, 3], [4, 5, 6]]
mb  = [[6, 3, 4], [5, 1, 2]]
mc  = [[0, 0, 0], [0, 0, 0]]

mat_add(ma, mb, mc)

for i in range(0, 2, 1):
    for j in range(0, 3, 1):
        print "%3d" % mc[i][j],
    print
N:\>python lesson_06_051.py
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Ruby

更新日 : 2010.11.01
def mat_add(a, b, c)
    for i in 0..1
        for j in 0..2
            c[i][j] = a[i][j] + b[i][j]
        end
    end
end

ma  = [[1, 2, 3], [4, 5, 6]]
mb  = [[6, 3, 4], [5, 1, 2]]
mc  = [[], []]

mat_add(ma, mb, mc)

for i in 0..1
    for j in 0..2
        printf("%3d", mc[i][j])
    end
    puts ""
end
L:\>ruby  l:\lesson_06_051.rb
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PowerShell

Scala

F#

C

更新日 : 2010.10.08
#include <stdio.h>

void mat_add(const int ma[][3], const int mb[][3], int mc[][3])
{
    int i, j;
    for (i = 0; i < 2; i++)
        for (j = 0; j < 3; j++)
            mc[i][j] = ma[i][j] + mb[i][j];
}

int main(int argc, char* argv[])
{
    int ma[2][3] = {{1, 2, 3}, {4, 5, 6}};
    int mb[2][3] = {{6, 3, 4}, {5, 1, 2}};
    int mc[2][3] = {0};

    mat_add(ma, mb, mc);

    int i, j;
    for (i = 0; i < 2; i++)
    {
        for (j = 0; j < 3; j++)
            printf("%3d", mc[i][j]);
        putchar('\n');
    }

    return 0;
}
R:\>lesson051\project1.exe
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C++

C++Builder

VC++

C#

Java

更新日 : 2010.11.05
class Lesson051 {
    public static void main(String[] args) {
        int ma[][] = {{1, 2, 3}, {4, 5, 6}};
        int mb[][] = {{6, 3, 4}, {5, 1, 2}};
        int mc[][] = new int[2][3];

        mat_add(ma, mb, mc);

        for (int i = 0; i < 2; i++)
        {
            for (int j = 0; j < 3; j++)
                System.out.printf("%3d", mc[i][j]);
            System.out.print('\n');
        }
    }

    static void mat_add(final int ma[][], final int mb[][], int mc[][])
    {
        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 3; j++)
                mc[i][j] = ma[i][j] + mb[i][j];
    }

}
L:\>java Lesson051
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Objective-C

D

VB

VB.NET

Delphi

更新日 : 2010.09.24
program Project1;

{$APPTYPE CONSOLE}

uses
    SysUtils;

type
    TMyArray = array[1..2, 1..3] of Integer;

procedure mat_add(const ma:TMyArray; const mb:TMyArray; var mc:TMyArray);
var
    i, j: Integer;
begin
    for i := 1 to 2 do
        for j := 1 to 3 do
            mc[i, j] := ma[i, j] + mb[i, j];
end;

var
    ma: TMyArray = ((1, 2, 3), (4, 5, 6));
    mb: TMyArray = ((6, 3, 4), (5, 1, 2));
    mc: TMyArray;
    i, j: Integer;
begin
    mat_add(ma, mb, mc);

    for i := 1 to 2 do
    begin
        for j := 1 to 3 do
            write(format('%3d', [mc[i, j]]));
        writeln('');
    end;
end.
S:\>lesson051\project1.exe
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Ada

PL/SQL

T-SQL

関数型

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