在C中实现的链接列表中使用单指针和双指针 [英] Using single versus double pointers in Linked lists implemented in C

问题描述

我正在编写这段代码，以便在链表的末尾添加元素:

I was writing this code for adding element at the end of linked list:

struct node{
    int info;
    struct node* link;
};

void append ( struct node **q, int num )  
{

struct node *temp, *r ;

if ( *q == NULL )       // if the list is empty, create first node
{
    temp = (struct node*) malloc ( sizeof ( struct node ) ) ;
    temp -> info = num ;
    temp -> link = NULL ;
    *q = temp ;        
}
else{
    temp = *q ;         

    /* go to last node */
    while ( temp -> link != NULL )
        temp = temp -> link ;

    /* add node at the end */
    r = (struct node *)malloc ( sizeof ( struct node ) ) ;
    r -> info = num ;
    r -> link = NULL ;
    temp -> link = r ;
}
}

，我这样调用append函数: append(&list, 10);其中list是指向链表的指针

and I call append function like this: append(&list, 10); where list is the pointer to the linked list

此代码有效，但是如果我在附加函数中使用单指针(使用* q代替** q)并进行相应的更改(如下所述，也称为调用")不起作用.下面的代码有什么问题?:

This code works, but if I use single pointer in append function(using *q instead of **q) and make changes accordingly (as done below and also when I call it), it doesn't work. What is wrong with the code below?:

void append ( struct node *q, int num )  
{

struct node *temp, *r ;

if ( q == NULL )       // if the list is empty, create first node
{
    temp = (struct node*) malloc ( sizeof ( struct node ) ) ;
    temp -> info = num ;
    temp -> link = NULL ;
    q = temp ;        
}
else{
    temp = q ;         

    /* go to last node */
    while ( temp -> link != NULL )
        temp = temp -> link ;

    /* add node at the end */
    r = (struct node *)malloc ( sizeof ( struct node ) ) ;
    r -> info = num ;
    r -> link = NULL ;
    temp -> link = r ;
}
}

推荐答案

因为在第二个示例中，q是调用者传递的指针的副本.调用者的原始指针永远不会被修改.

Because in the second example, q is a copy of the pointer passed in by the caller. The caller's original pointer never gets modified.

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