Question

我有一个矩阵（inA），其中包含0到23之间的值。

图像0的数据是：0,2,4,6，...

图像1的数据是：1,3,5,7 ......

我想通过每个图像将dat重新组织到另一个矩阵（ouB）。所以，ouB必须是：0,2,4,6,8,10,12,14,16 ......，1,3,5,7,9 ...

现在ouB是：0,2,4,6,1,3,5,7,8,10,12 ......

#include <stdio.h>
#include <stdlib.h>



int main(int argc, char * argv[]){

    int NbOfElmts = 4 , NbOfTech = 3 , NbOfImages = 2;

    int *inA =  malloc( NbOfElmts * NbOfTech * NbOfImages * sizeof( *inA ) );
    int *ouB =  malloc( NbOfElmts * NbOfTech * NbOfImages * sizeof( *ouB )  );


    //fill inA
    for ( int i = 0; i <  NbOfTech; i++ )
    {
        for ( int j = 0; j < NbOfElmts; j++ )
        {
            for ( int k = 0; k < NbOfImages; k++ )
            {
                inA[ i * NbOfElmts * NbOfImages + j * NbOfImages + k ] = i * NbOfElmts * NbOfImages + j * NbOfImages + k;
            }
            printf( "\n Data for #0 image: %d ",i * NbOfElmts * NbOfImages + j * NbOfImages );
            printf( "\n Data for #1 image: %d ",i * NbOfElmts * NbOfImages + j * NbOfImages + 1 );
        }

    }

    // print inA
    for ( int i = 0; i < NbOfTech; i++ )
    {
        for ( int j = 0; j < NbOfElmts; j++ )
        {
            for ( int k = 0; k < NbOfImages; k++ )
            {
                printf("\nA = %d",inA[ i * NbOfElmts * NbOfImages + j * NbOfImages + k ] );
            }
        }
    }


    printf("\n");
    // copy inA elements to ouB organized per Image
    for ( int i = 0; i <  NbOfTech; i++ )
    {
        for ( int j = 0; j < NbOfElmts; j++ )
        {
            for ( int k = 0; k < NbOfImages; k++ )
            {
                ouB[ i * NbOfElmts * NbOfImages + k * NbOfElmts + j ] = inA[ i * NbOfElmts * NbOfImages + j * NbOfImages + k ];
                //ouB[ i * NbOfElmts * NbOfImages + j * NbOfImages + k ] = inA[ i * NbOfElmts * NbOfImages + k * NbOfElmts + j ];
            }
        }
    }

    // print ouB
    for ( int i = 0 ; i < NbOfElmts * NbOfTech * NbOfImages; i++ ) 
        printf( "\nB = %d",ouB[ i ] );



    printf("\n");
    free ( inA );
    free ( ouB );

    return 0;
}

整个问题在这里：

ouB[ i * NbOfElmts * NbOfImages + k * NbOfElmts + j ] = inA[ i * NbOfElmts * NbOfImages + j * NbOfImages + k ];

Answer 1

尝试：

ouB[ k*NbOfElmts*NbOfTech + i * NbOfElmts  + j ] = inA[ i * NbOfElmts * NbOfImages + j * NbOfImages + k ];

而不是：

ouB[ i * NbOfElmts * NbOfImages + k * NbOfElmts + j ] = inA[ i * NbOfElmts * NbOfImages + j * NbOfImages + k

每个图像的数据重组

1 个答案: