Question

#include <stdio.h>
#include <stdlib.h>
struct node{
int data;
    struct node* next;
};

struct node* nc(int data) {

    struct node* temp = (struct node*)(malloc(sizeof(struct node)));
    temp->data = data;
    temp->next = NULL;
    return temp;
}
int getcount(struct node* head){
    int count = 0;
    struct node* temp= head;
    while(temp!=NULL){
        count++;
        temp=temp->next;
    }
    return count;
}

int getpositionofelement(int data,struct node* head){
    int count=0;
    while(head->data!=data){
        count++;
        head=head->next;
    }
    return count;
}
int main()
{
    struct node h;
    struct node* head = &h; //(struct node*)malloc(sizeof(struct node));
    h.data=5;
    struct node* a = nc(19);
    struct node* b = nc(25);
    struct node* c = nc(12);
    h.next = a;
    a->next=b;
    b->next=c;
    free(b);
    int count = getcount(head);
    printf("the count is %d \n",count);
    int l = 5;
    printf("the position of %d in linkedlist is %d \n",l,getpositionofelement(l,head));
    while(head!=NULL){
        printf("%d \n",head->data);
        head=head->next;    
    }
}

当我释放节点b时，为什么链接列表不会在b终止？将指针作为参数后，free()函数如何工作？

Answer 1

free（）从堆中释放内存，这就是全部。您需要将指针设置为null。你的指针可以包含一个存储器的地址，这个地址是免费的：它这是一个无效的地址，但它仍可存储在您的指针变量中。

free（）函数如何工作？

1 个答案: