Question

I am trying to study C, and I am running in to troubles using char* and char arrays. I am using a generic hash-set container from a library (which I don’t want to describe in details). This library includes the function

void *HashSetLookup(hashset *h, const void *elemAddr);

which I have to use to search in the hash set to see if the element already exists there (the hash and compare functions are part of the hashset struct). In this case I use the hashset to store pointers to C-strings, or more specifically (char * *) . My problem is that the following code gives a segmentation fault:

    char word[1024];
    /* Some code that writes to the word buffer */
    HashSetLookup(stopList, &word);

while this code works fine (and as expected):

    char word[1024];
    /* The same code as before that writes to the word buffer */
    char* tmp = strdup(word);
    HashSetLookup(stopList, &tmp);
    free(tmp);

I thought char word[] and char* were basically the same thing. The only difference being that char word[1024] is in the stack with a fixed length of 1024, but tmp in the heap occupying only as much space as necessary (strlen(word)+1).

Therefore I don’t understand why I have to make a copy of the string in the heap to be able to call this function. Why does this happen? Is there some more fundamental difference between char* tmp = strdup(“something”) and char word[1024] = “something”?

Answer 1

You mention you need a char ** and there lies the problem: for an array, word and &word mean the same thing – the actual location of the array contents. The reason it works when you use a pointer is because the “pointer” is stored at a different location, while it points to the same array. You don’t need an strdup, you simply need to create a pointer:

char* tmp = word;
HashSetLookup(stopList, &tmp);