设置一个指针，其余为NULL

Question

I'm trying to implement a linked list and I have come upon a very stubborn and frustrating problem. Here is the relevant code.

Node* current = list->head;
Node* previous = malloc(sizeof(Node*));
previous = NULL;
while(current != NULL){
    if(current == NULL){
        printf("current is null");
    }
    last = current;
    current = current->next;
    if(last == NULL){
        printf("last is null");
    }
} 

Now the problem with it is it is printing

"last is null"

frequently but doesn't print out

"current is null"

once. If current is not null and i set last = current then why does last stay null?

Thank you for any and all insights

Answer 1

Firstly, you don't want to use sizeof(Node*) which will allocate the size of a pointer (32-bit / 64-bit / something else depending on the platform). It is pointless since you can and already do define a pointer using Node* previous .

Secondly, expression in while() checks if current is not equal to NULL . If current is equal to NULL it will simply exit the loop so your if(current == NULL) comparison won't execute.

You can make that comparison valid if you put if(current == NULL) after current = current->next; . However, if last is null is printed, something else is also completely flawed with your code because current has to be null before last . It is surprising it is not giving a segmentation fault or something.