将malloc用于2D数组时出现分段错误

Question

I create a 2-D array using malloc . When I use printf to print the array element in for loop, everything is fine. But when I want to use printf in main, these is a Segmentation fault: 11.

Could you please tell me what the problem with the following code is?

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

void initCache(int **cache, int s, int E){
int i, j;
/* allocate memory to cache */
cache = (int **)malloc(s * sizeof(int *)); //set 
for (i = 0; i < s; i++){
    cache[i] = (int *)malloc(E * sizeof(int)); //int

    for(j = 0; j < E; j++){
        cache[i][j]  = i + j;   
        printf("%d\n", cache[i][j]);
    }
  }
}


main()
{
    int **c;

    initCache (c, 2, 2);

    printf("%d\n", c[1][1]);  // <<<<<<<<<< here

}

Answer 1

Since your cache is a 2D array, it's int** . To set it in a function, pass int*** , not int** . Otherwise, changes to cache made inside initCache have no effect on the value of c from main() .

void initCache(int ***cache, int s, int E) {
    int i, j;
    /* allocate memory to cache */
    *cache = (int **)malloc(s * sizeof(int *)); //set 
    for (i = 0; i < s; i++) {
        (*cache)[i] = (int *)malloc(E * sizeof(int)); //int
        for(j = 0; j < E; j++){
            (*cache)[i][j]  = i + j;   
            printf("%d\n", (*cache)[i][j]);
        }
    }
}

Now you can call it like this:

initCache (&c, 2, 2);

Answer 2

You changed a local variable, which won't effect the local variable c in main.

If you want to allocate in the function, why pass a variable? Return it from the function.

int **c = initCache(2, 2);

Answer 3

You could use a return , or else a *** as suggested by others. I'll describe the return method here.

initCache is creating and initializing a suitable array, but it is not returning it. cache is a local variable pointing to the data. There are two ways to make this information available to the calling function. Either return it, or pass in an int*** and use that to record the pointer value.

I suggest this:

int** initCache(int **cache, int s, int E){
   ....
   return cache;
}


main()
{
   int **c;
   c = initCache (2, 2);
   printf("%d\n", c[1][1]);   <<<<<<<<<< here
}

====

Finally, it's very important to get in the habit of checking for errors. For example, malloc will return NULL if it has run out of memory. Also, you might accidentally as for a negative amount of memory (if s is negative). Therefore I would do:

cache = (int **)malloc(s * sizeof(int *));
assert(cache);

This will end the program if the malloc fails, and tell you what line has failed. Some people (including me!) would disapprove slightly of using assert like this. But we'd all agree it's better than having no error checking whatsoever!

You might need to #include <assert.h> to make this work.