释放的指针未分配/ Segfault

Question

I thought I understood how dynamic memory worked basically, but I did the following. I use this function to allocate dynamic memory:

int get_alota_mem(char*** const ptr)
{
    int result = EXIT_SUCCESS;
    bool success = true;

    if (success && (*ptr = calloc(10, sizeof **ptr)) == NULL)
    {
        result = EXIT_FAILURE;
        success = false;
    }

    for (int i = 0; i < 10; i++)
    {
        if (success && ((*ptr)[i] = calloc(20, sizeof ***ptr)) == NULL)
        {
            result = EXIT_FAILURE;
            success = false;
        }
    }

    return result;
}

I use this function to free the allocated dynamic memory:

void free_alota_mem(char** const ptr)
{
    for (int i = 0; i < 10; i++)
    {
        free(ptr[i]);  // segfault!
        ptr[i] = NULL;
    }

    free(ptr);
}

In between those two calls I can use the memory and don't encounter any segfaults. But when I try to free the memory, I get a segfault with the first value I try to free.

I know I could just do char* ptr[10] and just use an auto allocated array instead of a dynamically allocated pointer, which would save me some hassle, but I wanted to try and make this work.

EDIT: My main:

int main(void)
{
    char** mem = NULL;
    get_alota_mem(&mem);
    mem[0] = "Hello";
    mem[1] = "world";
    free_alota_mem(mem);
}

Answer 1

You tried to free string literals. Passing pointer that is not NULL and not what is allocated via memory management functions such as malloc() to free() will invoke undefined behavior .

To copy strings, use strcpy() function.

#include <string.h>

int main(void)
{
    char** mem = NULL;
    get_alota_mem(&mem);
    strcpy(mem[0], "Hello");
    strcpy(mem[1], "world");
    free_alota_mem(mem);
}