链表头双指针传递 [英] Linked list head double pointer passing

问题描述

我在一些书/教程看到了这一点。

I have seen this in some book/ tutorial.

当你的头指针（链表）到一个函数传递，你需要把它作为一个双指针。

When you pass in the head pointer (of linked list) into a function, you need to pass it as a double pointer.

有关如：
//这是反向链接的列表，其中头指向第一个节点。

For eg: // This is to reverse a linked list where head points to first node.

void nReverse(digit **head)
{
    digit *prev=NULL;
    digit *curr=*head;
    digit *next;

    while(curr!=NULL)
    {
        next=curr->next;
        curr->next=prev;
        prev=curr;
        curr=next;
    }
    *head=prev;
    return;
}

这工作得很好。

它也可以当我使用单指针一样，

It also works when I use single pointer like,

void nReverse(digit *head)
{
    digit *prev=NULL;
    digit *curr=head;
    digit *next;

    while(curr!=NULL)
    {
        next=curr->next;
        curr->next=prev;
        prev=curr;
        curr=next;
    }
    head=prev;
    return;
}

我试图通过使用头指针打印列表。无论是功能正常工作。

I tried printing the list by using the head pointer. Both the functions work fine.

我缺少的东西吗？

谢谢，

推荐答案

这是非常类似于C code，而不是C ++。

This is very C-like code, not C++.

基本上，当事情是​​按值传递的功能上的数据的一个副本操作

Basically, when something is passed by-value the function operates on a copy of the data:

void foo(int i)
{
    i = 5; // copy is set to 5
}

int x = 7;
foo(x);
// x is still 7

在C，代替你传递一个指针变量，可以改变这种方式：

In C, you instead pass a pointer to the variable, and can change it that way:

void foo(int* i)
{
    *i = 5; // whatever i points to is set to 5
}

int x = 7;
foo(&x);
// x is 5

有关你，而不是一个 INT 这是一个位* 。 （在一个指针指向由此而来。）

For you, instead of an int it's a digit*. (Resulting in a pointer to pointer.)

在C ++中，引入了引用。引用是一个别名到另一个对象。所以，你会做这样的事情：

In C++, references were introduced. A reference is an alias to another object. So you'd do something like this:

void foo(int& i) // i is an alias to another value
{
    i = 5; // x is set to 5
}

int x = 7;
foo(x); // pass x as alias, not address of x.
// x is 5

一个参考一般preferred，因为它强制执行，你实际上是指一个对象，并简化了电话和工作code。

A reference is generally preferred, since it enforces that you actually refer to an object, and simplifies both calling and operating code.

当然在C ++中，你会不会实现一个列表你自己，你会使用的std ::列表。

Of course in C++ you wouldn't implement a list yourself, you'd use std::list.

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