The simple transition of matrix but what went wrong? - c

Question

#include <stdio.h>
main(){
int A[3][2] = {0} ;
printf("A = \n");
for(int x = 0 ; x < 3 ; x++){
for (int y = 0 ; y < 2 ; y ++){
A[x][y] = (x+1)*1 + (x*1+3)*y ;
printf("A[%d][%d] = %d ", x , y ,A[x][y]);
}
printf("\n");
}
printf("\nAT = \n");
for (int p = 0 ; p < 2 ; p++){
for (int q = 0 ; q < 3 ; q++){
A[p][q] = A[q][p];
printf("A[%d][%d] = %d ", p ,q , A[p][q]);
}
printf("\n");
}
}
Why does AT[1][0] = 2 not 4?
I have tested it for an hour for this simple question, but I have no idea.

Answer 1

If you transpose a non-square matrix then the dimensions of the resulting matrix will also be transposed. Not only that, if you were going to transpose in-place you would need to swap pairs of elements, otherwise you would be overwriting some of the elements before they had been transposed.
So, to solve both these problems you should probably just use a second matrix with the correct dimensions for the transposed result, e.g.
#include <stdio.h>
int main()
{
int A[3][2] = {0};
int AT[2][3] = {0}; // <<<
printf("A = \n");
for(int x = 0 ; x < 3 ; x++){
for (int y = 0 ; y < 2 ; y ++){
A[x][y] = (x+1)*1 + (x*1+3)*y ;
printf("A[%d][%d] = %d ", x, y, A[x][y]);
}
printf("\n");
}
printf("\nAT = \n");
for (int p = 0 ; p < 2 ; p++){
for (int q = 0 ; q < 3 ; q++){
AT[p][q] = A[q][p]; // <<<
printf("A[%d][%d] = %d ", p, q, AT[p][q]);
}
printf("\n");
}
return 0;
}
LIVE DEMO

Answer 2

You are accessing the array out of bounds when the condition came,
A[0][2]=A[2][0]; // <--
Here You can access only until first location. So as like Paul R says Use the
separate array when you transpose the array.
For example consider your array is declared in memory like this.
Memory Address 125 126 127 128 129 130
^ ^ ^ ^ ^ ^
| | | | | |
Array accessing [0][0] [0][1] [1][0] [1][1] [2][0] [2][1]
what will happen when you are accessing the [0][2]. It will go to the [1][0].