Linked List head not being updated across function calls

问题

I trying to implement my own linked list and have been messing around with the code learning about dynamic memory allocation and pointers and such. When I try to add something to my linked list I get a segfault, and upon using the debugger I realized that it was because initially my linked list's head pointer was not pointing to null and then my add function was not recognizing the head as being empty. But I have an initialize function that is setting the linked list's head pointer to NULL but for some reason once I exit out of the initialize function and into the add function, the head is no longer pointing to NULL.

Here's my code:

list.h

typedef struct node{
    int value;
    struct node *next;
} Node;

typedef struct list{
    Node *head;
} List;

list.c

void initialize(List *l){
    if(l != NULL){
        l = malloc(sizeof(List));
        l->head = NULL;
    }
}

void add(List *l, int a){
    //Code
}

int main(){
    List l;
    initialize(&l)
    add(&l, 2);
}

As soon as I step into the add function and print out *l, I see that the head is not pointing to 0x0. And I've been scratching my head as to why it's not. I thought it was something to do with pass by value but I don't think it is. What am I doing wrong here?

回答1:

Yes, pass-by-value is your culprit. You are passing a pointer by value.

Suppose l in your main() is at address 0xABCD. Then your main() gets compiled to

int main(void) {
    List l;
    initialize(0xABCD);
    add(0xABCD, 2);
}

and your initialize() call looks like this (suppose malloc() succeeds and allocates memory at address 0xCDEF:

void initialize(List *l) {
    if(l != 0x0) {
        l = 0xCDEF;    // malloc()
        l->head = 0x0;
    }
}

That l = 0xCDEF does not propagate to main(), because l was passed by value.

What you want to do is

void initialize(List **l) {
    if(l != NULL) {
        *l = malloc(sizeof(List)); // note dereferencing the passed-by-value pointer
        (*l)->head = NULL;
    }
}

int main(void) {
    List * l;
    initialize(&l);
    add(l, 2);
}

which will pass pointer to pointer to list (actually the address of the pointer in your main(). It allows the code in initialize() to change the l variable in main().

Alternatively, you can use

List * list_init() {
    List * retval = malloc(sizeof(List));
    if(retval == NULL) { // you should check malloc return value
        // abort(), print warning or just
        return NULL;
    }
    retval->head = NULL;
    return retval;
}

int main(void) {
    List * l = list_init();
    if(l == NULL) {
        // handle the error
    }
    add(l, 2);
}

回答2:

You declare a List in main() that lives on the stack. You pass a pointer to that List to initailize(). You then create a new List on the heap. When you return from initialize() you still are using the List on the stack that you had in the beginning. The List on the heap is leaked and you cannot access it. So you never initialized the List you pass as a pointer to add(). You can forget about initialize() and just have

l.head = NULL;

instead.

回答3:

Did you code compile this line l->malloc(sizeof(list)); seems odd.

Create a structure with only one argument is not really useful, a simple typedef should do the job : typedef Node* List

来源：https://stackoverflow.com/questions/36817825/linked-list-head-not-being-updated-across-function-calls

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linked-list