C中的指针N链表 [英] pointers N linked list in C

问题描述

我写了以下无效的代码.应用程序在打印方法上崩溃.知道哪里出了问题吗?

I wrote to following code which does not work. The application crashes on the print method. Any idea where it goes wrong?

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

typedef struct
{
    int item;
    struct LinkedList *Next;
} LinkedList;

int main()
{
    LinkedList vaz;
    vaz.item = 14123;
    vaz.Next = 0;
    add(&vaz,123);
    add(&vaz,9223);
    Print(&vaz);
    return 0;
}
void Print(LinkedList* Val)
{
    if(Val != 0){
        printf("%i\n",(*Val).item);
        Print((*Val).Next);
    }
}
void add(LinkedList *Head, int value)
{
    if((*Head).Next == 0 )
    {
        LinkedList sab;
        sab.item = value;
        sab.Next = 0;
        (*Head).Next = &sab;
    }
    else{
       add((*Head).Next,value);
    }
}

推荐答案

您的add方法应该动态分配内存，您使用的是自动分配(局部变量)，当您离开"时会释放该分配功能.当您稍后尝试访问此内存时，这将导致未定义的行为.

Your add method should allocate memory dynamically, you are using automatic allocation (local variables), which is freed when you 'leave' the function. This will cause an undefined behavior when you try to access this memory later on.

void add(LinkedList *Head, int value)
{
    if((*Head).Next == 0 )
    {
        LinkedList sab; <-- need to be allocated dynamically
        sab.item = value;
        sab.Next = 0;
        (*Head).Next = &sab;
    }
    else{
       add((*Head).Next,value);
    }
}

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