free char*: invalid next size (fast) [duplicate]

Question

This question already has an answer here:

• Facing an error “*** glibc detected *** free(): invalid next size (fast)” 2 answers

I am freeing a char* after a concatenation process.
But I receive this error:

free(): invalid next size (fast): 0x0000000001b86170

Below is my code:

void concat(stringList *list) {
    char *res = (char*)malloc(sizeof(char*));

    strcpy(res, list->head->string);

    list->tmp = list->head->next;
    while (list->tmp != NULL) {
        strcat(res, ",");
        strcat(res, list->tmp->string);
        list->tmp = list->tmp->next;
    }

    printf("%s\n", res);

    free(res);
}

Answer 1

Your code is wrong.

You are allocating space for a single pointer (malloc(sizeof(char*))), but no characters. You are overwriting your allocated space with all the strings, causing undefined behavior (in tihs particular case, corrupting malloc()'s book-keeping data).

You don't need to allocate space for the pointer (res), it's a local variable. You must allocate space for all the characters you wish to store at the address held by the pointer.

Since you're going to be traversing a list to find strings to concatenate, you can't know the total size upfront. You're going to have to do two passes over the list: one to sum the strlen() of each string, then allocate that plus space for the separator and terminator, then another pass when you actually do the concatenenation.