Question

I don’t understand why, in this code, the call to “free” cause a segmentation fault:

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

char *char_arr_allocator(int length);

int main(int argc, char* argv[0]){

    char* stringa =  NULL;
    stringa = char_arr_allocator(100);  
    printf("stringa address: %p\n", stringa); // same address as "arr"
    printf("stringa: %s\n",stringa);
    //free(stringa);

    return 0;
}

char *char_arr_allocator(int length) {
    char *arr;
    arr = malloc(length*sizeof(char));
    arr = "xxxxxxx";
    printf("arr address: %p\n", arr); // same address as "stringa"
    return arr;
}

Can someone explain it to me?

Thanks,
Segolas

Answer 1

You are allocating the memory using malloc correctly:

arr = malloc(length*sizeof(char));

then you do this:

arr = "xxxxxxx";

this will cause arr point to the address of the string literal "xxxxxxx", leaking your malloced memory. And also calling free on address of string literal leads to undefined behavior.

If you want to copy the string into the allocated memory use strcpy as:

strcpy(arr,"xxxxxxx");