为何以及何时需要双指针？ [英] why and when is a double-pointer required?

问题描述

我一直是一个系统管理员，我大部分的生活，但最近决定练习一些我开发的知识和尝试DEVOPS位置。我有这样练了一些C和Python技能，写了一些code插入一个数字，一个链表。

I have been a sysadmin most of my life, but recently decided to practice some my dev knowledge and attempt a devops position. I have as such been practicing some C and Python skills and wrote some code for inserting a number into a linked list.

void list_insert(struct list *h, int d)
{
        struct list *elem = malloc(sizeof(struct list));

        elem->data = d;
        elem->next = NULL;

        if (!h) {
                h = elem;
        } else {
                elem->next = h;
                h = elem;
        }
}           

我注意到，这个功能似乎并没有改变变量h外视图（即无论是传递给list_insert），我注意到，在插入函数结束打印似乎工作。所以有试图寻找答案在网上，我无法找到任何东西，很明显，但我发现大多数列表的实现将有双指针。我改变了我的功能，采用双指针，然后它突然开始工作。可能有人帮助我了解这里发生了什么，因为我知道指针管理是一个重要的概念，我想我明白一个指针是什么，以及它如何与记忆，但我不认为我明白为什么一个指针不得到改变，而双指针一样。

I noticed that this function doesn't seem to alter the outside view of the variable h (i.e whatever is passed to list_insert) and I noticed that printing at the end of the insert function seems to work. So having tried to look for answers online, I couldn't find anything, obvious but I found most list implementations would have double-pointers. I changed my function to use double-pointers and then it suddenly started working. Could someone help me understand what's happening here as I know pointer management is an important concept and I think I understand what a pointer is and how it relates to memory, but I don't think I get why a single pointer does not get changed, whereas a double-pointer does.

谢谢！

推荐答案

在C，参数的功能是由值传递。即使指针由值传递。

In C, arguments to function are passed by values. Even pointers are passed by values.

例如：

#include<malloc.h>
#include<stdio.h>
int allocatingMemory(int* ptr)
{
    ptr = malloc(sizeof(int)); 
    if(ptr==NULL)
        return -1;
    else 
        return 0;
}// We are not returning the pointer to allocated memory

int main(void)
{
    int* ptr;
    int allocated = allocatingMemory(ptr);
    if(allocated == 0)
    {
        *ptr = 999;// Boom!!! 
        free(ptr);
    }

    return 0;
}

要解决这个问题，我们使用

To overcome this issue, we use

int allocatingMemory(int** ptr)
{
    *ptr = malloc(sizeof(int));
    if(*ptr == NULL)
        return -1;
    else 
        return 0;
}

int main(void)
{
    int* ptr;
    int isAllocated = allocatingMemory(&ptr);
    if(isAllocated == 0)
    {
        *ptr = 999;
        free(ptr);
    }

    return 0;
}

如果您正在使用链表 s工作，并比方说，你要修改的头。您将通过 A 指针指针（需要注意的是，它不称为双指针）头节点。

If you are working with linked lists and say for example, you want to modify the head. You will pass a pointer to pointer (Note that, it is not called as double pointer) to head node.

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