该程序不打印任何内容。我不明白它的行为。 [英] This program doesn't print anything. I don't understand it's behaviour.

问题描述

  #include    <   stdio.h  >  
   #include   <   conio .h  >  
   #include   <   alloc.h  > ;  
   struct 已链接
 { int 数据; 
  struct  linked * next; 
}; 
  void  insert（ struct  linked * root， int  ele）; 
 
  void  main（）
 {
  struct  linked * root = NULL; 
  struct  linked * temp; 
 insert（root， 4 ）; 
 insert（root， 6 ）; 
 insert（root， 7 ）; 
 insert（root， 1 ）; 
 insert（root， 3 ）; 
 insert（root， 9 ）; 
 temp = root; 
  while （temp！= NULL）
 {
 printf（ ％d \ n，temp-> data）; 
 temp = temp-> next; 
} 
} 
 
 
  void  insert（ struct 链接* root， int  ele）
 {
  struct 已链接* newnode =（ struct  linked *）malloc（ sizeof （ struct  linked））; 
 newnode-> data = ele; 
 newnode-> next = root; 
 root = newnode; 
} 

解决方案

该程序表现得像预期的那样。

  void  insert（ struct  linked * root， int  ele）; 

永远不会改变root。



问题是在你的声明中插入到链表中，你传递的是指针的值，而不是指针的地址。这意味着指针的副本被传递到堆栈而不是指针的地址。



将函数声明为：

< pre lang =c ++> void insert（ struct linked ** root， int ele）;

然后电话将采用以下形式：

 insert（& root， 6 ）; 

如果需要，请尝试并寻求更多帮助。





你需要在脑海中清醒关于指针本身与它指向的内容之间的区别。

在C ++中，您可以使用引用并通过声明修复程序：

  void  insert（ struct  linked *& root， int  ele）; 

见这里（参考文献1总结它up）：

http://www.eskimo.com/~scs /cclass/int/sx8.html [ ^ ]



指针指针和指针引用 [ ^ ]



因为SA说调试器是你的朋友有这种问题

#include<stdio.h>
#include<conio.h>
#include<alloc.h>
struct linked
{int data;
struct linked *next;
};
void insert(struct linked *root,int ele);

void main()
{
struct linked *root=NULL;
struct linked *temp;
insert(root,4);
insert(root,6);
insert(root,7);
insert(root,1);
insert(root,3);
insert(root,9);
temp=root;
while(temp!=NULL)
{
printf("%d\n",temp->data);
temp=temp->next;
}
}


void insert(struct linked *root,int ele)
{
struct linked *newnode=(struct linked *)malloc(sizeof(struct linked));
newnode->data=ele;
newnode->next=root;
root=newnode;
}

解决方案

The program is behaving as one would expect.

void insert(struct linked *root,int ele);

never changes root.

The problem is that in your declaration for an insert into the linked list you are passing the value of a pointer and not the address of the pointer. This means a copy of the pointer is passed onto the stack instead of the address of the pointer.

Declare the function as:

void insert(struct linked **root,int ele);

Then calls will be of the form:

insert(&root,6);

Try that and ask for more help if required.

[EDIT]
You need to be clear in your mind about the difference between the pointer itself and what it points to.
In C++ you could have used a reference and fixed the program by declaring:

void insert(struct linked *&root,int ele);

See here (Reference 1 sums it up):
http://www.eskimo.com/~scs/cclass/int/sx8.html[^]

Pointer to Pointer and Reference to Pointer[^]

As SA has said the debugger is your friend with this type of problem.

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