将结构的指针成员分配为null

Question

I am having some difficulty assigning a pointer inside a struct to null. Since pointers are pass by value, I can't find an easy way to do this. Maybe its been a long day and I can''t think straight. Anyways, here is my code:

void
init_wordrec (wordrec *rec)
{
    if ((rec = (wordrec *) malloc(sizeof(wordrec))) == NULL) {
        perror("Malloc failed");
        exit(1);
    }
    rec->word = NULL;
    rec->hits = 0;
    rec->nxt_wd = NULL;
}

Here is the wordrec struct:

typedef struct wordrec
{
    char *word;
    int hits;
    struct wordrec *nxt_wd;
} wordrec;

I want the actual word pointer to point to null, unfortunately my attempts have only caused gcc to complain loudly.

EDIT: Here is a method where I pass in word struct.

void
add_word (char *word, wordrec *head)
{
    wordrec *iter = head;
    wordrec *tmp;
    if (iter->word == NULL) { //This should be NULL but is not
        if ((iter->word = (char *) malloc((strlen(word) + 1) * sizeof(char))) == NULL) {
            perror("Malloc failed");
            exit(1);
        }
        strncpy(iter->word, word, strlen(word) + 1);
        iter->hits++;
        init_wordrec (iter->nxt_wd);
    } else if (strcmp(iter->word, word) < 0) {
        init_wordrec (tmp);
        if ((tmp->word = (char *) malloc((strlen(word) + 1) * sizeof(char))) == NULL) {
            perror("Malloc failed");
            exit(1);
        }
        strncpy(tmp->word, word, strlen(word) + 1);
        tmp->hits++;
        tmp->nxt_wd = head; 
        head = tmp;
    } else if (strcmp(iter->word, word) > 0) {
        add_word (word, iter->nxt_wd);
    } else {
        iter->hits++;
    }
}

Main:

int 
main()
{
    wordrec head;
    char word1[] = "Hello";
    init_wordrec (&head);
    add_word(word1, &head);
    return 0;
}

Answer 1

Did you mean something like this:

void init_wordrec(wordrec **rec)
{
    if ((*rec = (wordrec *) malloc(sizeof(wordrec))) == NULL) {
        perror("Malloc failed");
        exit(1);
    }
    (*rec)->word = NULL;
    (*rec)->hits = 0;
    (*rec)->nxt_wd = NULL;
}
....
wordrec *wr;
init_wordrec(&wr);

Answer 2

Assignment goes the other way — you assign a value, such as NULL , to a variable, such as rec->word .

You could pass a double pointer to the function:

void
init_wordrec (wordrec **rec)
{
    if ((*rec = (wordrec *) malloc(sizeof(wordrec))) == NULL) {
        perror("Malloc failed");
        exit(1);
    }
    (*rec)->word = NULL;
    (*rec)->hits = 0;
    (*rec)->nxt_wd = NULL;
}
/* ... */
wordrec *wr = 0;
init_wordrec(&wr);

but since you're allocating in the function (so you're just going to throw away the value passed in) making it return the new record usually simplifies the code:

wordrec*
init_wordrec(void)
{
    wordrec *rec = malloc(sizeof(*rec));
    if (rec == NULL) {
        perror("Malloc failed");
        exit(1);
    }
    rec->word = NULL;
    rec->hits = 0;
    rec->nxt_wd = NULL;
    return rec;
}
/* ... */
wordrec *wr = init_wordrec();

You shouldn't cast the result of malloc ; it will make the code compile even if you fail to include its prototype, but will fail badly at runtime.

Answer 3

A simpler option would be:

void init_wordrec (wordrec *rec)
{
    rec->word = NULL;
    rec->hits = 0;
    rec->nxt_wd = NULL;
}

which allows your code in main to remain unchanged:

int main()
{
    wordrec head;
    char word1[] = "Hello";
    init_wordrec (&head);

Another advantage of this is that users of wordrec have the option to use automatic allocation or dynamic allocation (whereas the other proposed answers force dynamic allocation to be used).