用C转置矩阵函数，函数接收和返回2D数组

Question

I need to create a function that takes a matrix and returns it transpose. The only requirement is that it directly returns a matrix, not just modifies it by reference. Here's what I've done so far:

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

#define ROW 100000000
#define COL 100000000

int (*(f_MatTrans)(int mat[][COL], int r, int c))[COL];

int main(void)
{
int x[2][2]={1,2,3,4};
int (*a)[2];
a=f_MatTrans(x,2,2);

for(int i=0; i<2; i++)
{
    for(int j=0; j<2; j++)
    {
        printf("X[%d][%d]=%d\n",i,j,x[i][j]);
        printf("A[%d][%d]=%d\n",i,j,a[i][j]);
    }
}

return 0;
}

int (*(f_MatTrans)(int mat[][COL], int r, int c))[COL]
{
int a[c][r];

for(int i=0; i<r; i++)
{
    for(int j=0; j<c; j++)
    {
        a[j][i]=mat[i][j];
    }
}

return a;

}

The purpose of this is to include the function on a library created by myself, just in case it is useful information.

Answer 1

You cannot return a pointer to the local array, because that ceases to exist when the function returns. If you want your function to create the result array (not write to some other array that is passed into the function), you must use malloc() in these cases:

//The return type is actually `int (*)[r]`, but C doesn't like that.
int* f_MatTrans(int r, int c, int mat[][c]) {
    int (*a)[r] = malloc(c*sizeof(*a));
    for(int i=0; i<r; i++) {
        for(int j=0; j<c; j++) {
            a[j][i]=mat[i][j];
        }
    }
    return *a;
}

Note that I changed the array types: If you declare mat as int mat[][COL] , the number COL will be used to calculate the offset mat[1][0] , which will be 100000000 integers after the first element in your case, while the array that you pass in only contains four integers. This is undefined behavior, and your program is allowed to format your harddrive if you do this.

Unfortunately, it is not possible for the type of the returned pointer to depend on the value of an argument to the function. That is why I changed the return type to a plain integer pointer, you must document that this is meant to be a pointer of type int (*)[r] .

You would use the function above like this:

int main(void) {
    int x[2][3]={1,2,3,4,5,6};
    int (*a)[2] = (int (*)[2])f_MatTrans(2, 3, x);

    for(int i=0; i<2; i++) {
        for(int j=0; j<2; j++) {
            printf("X[%d][%d]=%d\n",i,j,x[i][j]);
            printf("A[%d][%d]=%d\n",i,j,a[i][j]);
        }
    }

    free(a);    //Cleanup!
    return 0;
}

Answer 2

The code in the question (when I read it) doesn't compile because the array x is not compatible with the function signature.

I'm not clear what the real constraints on your problem are. The easy way to do it in C99 or C11 is with VLA notation:

#include <stdio.h>

static void MatrixTranspose(int r, int c, int src[r][c], int dst[c][r])
{
    for (int i = 0; i < r; i++)
        for (int j = 0; j < c; j++)
            dst[j][i] = src[i][j];
}

int main(void)
{
    int x[3][2] = { { 0, 1 }, { 2, 3 }, { 4, 5 } };
    int y[2][3];

    MatrixTranspose(3, 2, x, y);

    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 2; j++)
        {
            printf("X[%d][%d]=%d  ", i, j, x[i][j]);
            printf("Y[%d][%d]=%d\n", j, i, y[j][i]);
        }
    }

    return 0;
}

Sample output:

X[0][0]=0  Y[0][0]=0
X[0][1]=1  Y[1][0]=1
X[1][0]=2  Y[0][1]=2
X[1][1]=3  Y[1][1]=3
X[2][0]=4  Y[0][2]=4
X[2][1]=5  Y[1][2]=5

My suspicion is that you are supposed to be doing something different (notationally more complex), but it is not yet clear what.