How to calculate indices for multiple sub-indices from a single for-loop?-CodePudding

Let's say that instead of writing

for (int i = 0; i < I;   i) {
    for (int j = 0; j < J;   j) {
        /* do whatever */
    }
}

I write

for (int k = 0; k < I * J;   k) {
 /* who do I calculate indices i and j? */
}

how do I calculate the two previous indices i and j?

CodePudding user response：

Here you are

i = k / J;
j = k % J;

Here is a demonstration program

#include <stdio.h>

int main( void )
{
    enum { I = 2, J = 3 };
    int a[I][J] =
    {
        { 1, 2, 3 },
        { 4, 5, 6 }
    };

    for ( int i = 0; i < I; i   )
    {
        for ( int j = 0; j < J; j   )
        {
            printf( "%d ", a[i][j] );
        }
        putchar( '\n' );
    }

    putchar( '\n' );
    
    for ( int k = 0, N = I * J; k < N;   k )
    {
        printf( "%d ", a[k / J][k % J] );
        if ( ( k   1 ) % J == 0 ) putchar( '\n' );
    }
}

The program output is

Or if to use a 3D array then the program can look the following way

#include <stdio.h>

int main()
{
    enum { I = 2, J = 3, K = 4 };
    int a[][J][K] =
    {
        {
            { 1,  2,  3,  4 },
            { 5,  6,  7,  8 },
            { 9, 10, 11, 12 }
        },

        {
            { 13, 14, 15, 16 },
            { 17, 18, 19 , 20 },
            { 21, 22, 23, 24 }
        }
    };

    for ( int i = 0; i < I; i   )
    {
        for ( int j = 0; j < J; j   )
        {
            for ( int k = 0; k < K;   k )
            {
                printf( "- ", a[i][j][k] );
            }
            putchar( '\n' );
        }
        putchar( '\n' );
    }

    putchar( '\n' );

    for ( int k = 0, N = I * J * K; k < N;   k )
    {
        printf( "- ", a[k / ( J * K )][k %  ( J * K ) / K][k % (J * K ) % K] );
        if ( ( k   1 ) % K == 0 ) putchar( '\n' );
        if ( ( k   1 ) % ( J * K ) == 0 ) putchar( '\n' );
    }
}

Its output is

 1  2  3  4 
 5  6  7  8 
 9 10 11 12 

13 14 15 16 
17 18 19 20 
21 22 23 24 


 1  2  3  4 
 5  6  7  8 
 9 10 11 12 

13 14 15 16 
17 18 19 20 
21 22 23 24

CodePudding user response：

#include <stdio.h>

int main(void) {
    const int I = 7;
    const int J = 4;
    
    for (int k = 0; k < I * J;   k) {
        int i = k/J;
        int j = k%J;
        
        printf("Calculated i,j: %d,%d\n", i, j);
    }
    return 0;
}

As the output demonstrates, the calculated i value goes from 0 to 6, and the calculated j value goes from 0 to 3.

Success #stdin #stdout 0.01s 5444KB
Calculated i,j: 0,0
Calculated i,j: 0,1
Calculated i,j: 0,2
Calculated i,j: 0,3
Calculated i,j: 1,0
Calculated i,j: 1,1
Calculated i,j: 1,2
Calculated i,j: 1,3
Calculated i,j: 2,0
Calculated i,j: 2,1
Calculated i,j: 2,2
Calculated i,j: 2,3
Calculated i,j: 3,0
Calculated i,j: 3,1
Calculated i,j: 3,2
Calculated i,j: 3,3
Calculated i,j: 4,0
Calculated i,j: 4,1
Calculated i,j: 4,2
Calculated i,j: 4,3
Calculated i,j: 5,0
Calculated i,j: 5,1
Calculated i,j: 5,2
Calculated i,j: 5,3
Calculated i,j: 6,0
Calculated i,j: 6,1
Calculated i,j: 6,2
Calculated i,j: 6,3