将动态分配的2d数组传递给函数

Question

The following code is not working correctly. I'm getting a segfault when I run the program. I ran my program through gdb and found out that the error is occuring in the fillArrays(int**,int) function.

GDB is displaying the following parameters for fillArrays(int**,int):

fillArrays (arrays=0x0,numArrays=3)

Here is the source code to my program

#include <stdlib.h> /* malloc and free */

#define MULTIPLIER          1
#define SMALL               10
#define BIG                 20

void allocateSmallArrays(int **arrays,int numArrays) {
    int index,freeIndex;
    int outerIndex,innerIndex;
    arrays = malloc(numArrays*sizeof(int*));
    if(arrays == NULL) {
        printf("out of memory\n");
        exit(1);
    }
    for(index = 0;index < numArrays;index++) {
        arrays[index] = malloc(SMALL*sizeof(int));
        if(arrays[index] == NULL) {
            printf("out of memory\n");
            exit(1);
        }
    }
}

void fillArrays(int **arrays,int numArrays) {
    int outerIndex,innerIndex;
    for(outerIndex = 0;outerIndex < numArrays;outerIndex++) {
        for(innerIndex = 0;innerIndex < SMALL;innerIndex++)
            arrays[outerIndex][innerIndex] = 0;
    }
}

void deallocateSmallArrays(int **arrays,int numArrays) {
    int index;
    for(index = 0;index < numArrays;index++)
        free(arrays[index]);
    free(arrays);
}

int main(void) {
   int numArrays  = (3 * MULTIPLIER);
   int **arrays = 0;

   allocateSmallArrays(arrays,numArrays);
   fillArrays(arrays,numArrays);
   deallocateSmallArrays(arrays,numArrays);

   arrays = 0;

   return 0;
}

I was under the assumption that since arrays was allocated in allocateSmallArrays, that passing it through fillArrays would 0 out the allocated arrays and then deallocate in the last function. How do I go about accomplishing this?

Answer 1

The problem is that allocateSmallArrays changes its own copy of the arrays pointer . So the result of the malloc is lost and after the function is done, in the caller arrays is still 0. You could:

• Pass a triple pointer int ***arrays and do to *arrays everything you're doing to arrays

• Return the pointer instead of void

A C FAQ deals with this very subject.