C编程错误，打印链接列表，在运行时执行的代码崩溃 [英] C Programming Error , printing linked list, at run time executed code crashes

问题描述

我正在处理链表和指针.这是一个包含推送功能的简单代码.在推送完我的元素并尝试打印第一个成员之后，执行的代码在运行时崩溃.但是，当将相同的指针传递给 print_list 函数并在print_list内应用 printf 函数时，它可以正常工作.但是，当直接在main函数中使用它并应用printf函数时，它会崩溃.

I'm working on linked lists and pointers. Here is a simple code including a push function. After pushing my elements, and trying to print the first member, the executed code crashes at run time. However, when passing the same pointer to the print_list function and applying the printf function inside print_list it works fine. But when using it directly in the main function and applying the printf function it crashes.

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


typedef struct list{
int order;
struct list *next;

}list;

void push(struct list **arg,int i);

int main()
{
struct list **ptr=NULL;

for(int i=0;i<10;++i){
    push(&ptr,i);
}
    print_list(&ptr); // Here works fine
    printf("%d\n", (*ptr)->order); //Here run time error
return 0;
}


void push(struct list **arg,int i){

   struct list *temp;
   temp= malloc(sizeof(list));

temp->order=i;

temp->next=*arg;

*arg=temp;

}


void print_list(list ** head) {


while ((*head) != NULL) {
    printf("%d\n", (*head)->order); //Here works fine too !
    *head = (*head)->next;
}
 }

推荐答案

这段代码中存在一些指针管理错误.

There are a couple pointer management errors in this piece of code.

void print_list(list ** head) {
  while ((*head) != NULL) {
    printf("%d\n", (*head)->order);
    *head = (*head)->next; // print_list is clearly a readonly function, you don't want to modify head of the list
  }
}

改为使用迭代器:

void print_list(list *head) { // Passing a copy of head pointer here, so real head can't be changed.
  struct list *iter = head;
  while (iter != NULL) {
    printf("%d\n", iter->order);
    iter = iter->next;
  }
}

struct list **ptr=NULL;-您想在此处声明指向列表头的指针，但是您正在声明指向该指针的指针.

struct list **ptr=NULL; - Your want to declare pointer to head of the list here, but you are declaring pointer to this pointer.

将其更改为:struct list *ptr = NULL;

此更改之后，您无需输入头像地址即可将其传递给print_list:

And after this changes you don't need to either take an address of head to pass it to print_list:

print_list(&ptr)-> print_list(ptr)

或在printf中取消引用:

or dereference it in printf:

printf("%d\n", (*ptr)->order)-> printf("%d\n", ptr->order)

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