How should I release a char pointer by passing a const char pointer for free?

Question

I have the code below in a main.c file:

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

struct mystruct {const char *name; int x;};

char *test_calloc_char_p(const char *char_p) {
  return (char *) calloc(1, (strlen(char_p) + 1) * sizeof(char));
}

int main(void) {
  struct mystruct *ms1;
  ms1 = (struct mystruct *) calloc(1, sizeof(struct mystruct));
  ms1->name = test_calloc_char_p("dlldffdl");
  ms1->x = 3;
  free(&ms1->name);
  free(ms1);
  return 0;
}

Which I compile with gcc main.c -o test and run with ./test, and I get:

free(): double free detected in tcache 2
Abortado

I don't understant, why? I should not release ms1->name? because when I comment this line I don't get the error, but why shouldn't I release something that I allocated?

Answer 1

This is incorrect:

free(&ms1->name);

The address of ms1->name is part of the allocated memory assigned to ms1. it is not the allocation for the name member.

You assigned allocated memory to ms1->name, so that is what you should pass to free:

free((char *)ms1->name);

Note that the cast is needed because otherwise you would be passing a const pointer to a function expecting a non-const pointer. The cast can be removed by changing the name member from const char * to char *.