如何将空指针（结构返回类型）更改为声明的结构？

Question

I am new to C programming. I am trying to implement Linked list by myself. I am encountering problem with pointers

I have function

void Insert(Node* head, int x)

to insert node at the beginning of Linked list. The problem is that when I insert the very first node and Node *head is NULL, the function Insert is not able to change the pointer address of null pointer to the newly created node. It seems as if the Node *head is passed by value and not by reference. Code is provided below. In order to debug how address is changed throughout the execution, I used printf function.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef struct Node {
    int data;
    struct Node *next;
} Node;

int main() {
    Node *head = (Node*) malloc(sizeof(Node));
    head = NULL;

    printf("head in main(): %d\n", head); // For example: 100

    Insert(head, 25);
    return 0;
}

void Insert(Node *head, int x) {
    Node *temp = (Node*) malloc(sizeof(Node));
    temp->data = x;
    temp->next = head;

    printf("temp->next address: %d\n", temp->next); // also 100

    head = temp;

    printf("%d\n", head); // not 100, something else i.e  200
}

Answer 1

It seems as if the Node *head is passed by value and not by reference.

That is exactly right -- in C every parameter is always passed by value. The pointer is a value, and that value is passed by value, and the call

Insert(head, 25);

can never change the value of the variable named head . It reads the value of the variable (this value is a null pointer), gives that value to the function and never touches the variable head again no matter what the function does.

(Note that in your program you have two variables that are both named head -- one in main() and the other in Insert() . The variable in Insert() silently disappears when the function returns; nothing will automatically try to copy its value to the similarly-named variable in main() ).

If you want to (conceptually) pass head by reference , you need to actually pass a pointer to it -- that is, in this case, a pointer to a pointer! You'd need to declare your function as

void Insert(Node **head, int x) { ... }

and call it as

Insert(&head, 25);

Then the actual parameter is a pointer to the variable head which gives the function a chance to update that variable, if you deference the parameter where appropriate:

// ...
temp->next = *head;
// ...
*head = temp;
// ...

Answer 2

Pass a pointer to a pointer to head. That way, you can set head to null.

void Insert(Node **head, int x) {

    ...

    if (*head == NULL) {
        *head = temp;
    }

    else {
        ...
        *head->next = temp;
    }
}

Usage:

Node *head = NULL;
Insert(&head, 10);

Answer 3

Having three answers suggesting the same I would like to offer an alternative: Instead of passing a Node ** to Insert() you could instead have it return the new head , thus:

Node *Insert( Node *head, int x )
{
     ... your code ...
     return head;
}

and if you call it by

head = Insert( head, 24 );

That is neither better nor worse then the other solution so you my do whatever you prefer

Answer 4

There are number of issues here.
1. Your printf statements need to be corrected.
2. To insert function you can pass double pointer.
3. Inside main function, you need not to do Node *head = (Node*) malloc(sizeof(Node));

I have modified your code as shown below. You can try running it and co-relate above points.

typedef struct Node {
    int data;
    struct Node *next;
} Node;

void Insert(Node **head, int x);

void Insert(Node **head, int x) {
    Node *temp = (Node*) malloc(sizeof(Node));
    temp->data = x;
    temp->next = *head;

    printf("temp->next address: %d\n", temp->data); // also 100

    *head = temp;

    printf("%d\n", (*head)->data); // not 100, something else i.e  200
}

int main() {
    Node *head = NULL;
    head = NULL;

    Insert(&head, 25);
    printf("head in main(): %d\n", head->data); // For example: 100
    return 0;
}