故事起源于Linus大神在回答水友提问的时候有这么一段: http://meta.slashdot.org/story/12/10/11/0030249/linus-torvalds-answers-your-questions

At the opposite end of the spectrum, I actually wish more people understood the really core low-level kind of coding. Not big, complex stuff like the lockless name lookup, but simply good use of pointers-to-pointers etc. For example, I’ve seen too many people who delete a singly-linked list entry by keeping track of the “prev” entry, and then to delete the entry, doing something like

if (prev)
    prev->next = entry->next;
else
    list_head = entry->next;

and whenever I see code like that, I just go “This person doesn’t understand pointers”. And it’s sadly quite common.

People who understand pointers just use a “pointer to the entry pointer”, and initialize that with the address of the list_head. And then as they traverse the list, they can remove the entry without using any conditionals, by just doing a “*pp = entry->next”.

大意是说，当从一个单向链表中删除一个元素时，很多人会使用一个prev指针来记录被删除的那个节点的前一个元素的位置，然后使用

1

prev->next = entry->next
的方式去删除。这样的人简直弱爆了，根本就不懂得指针。

可能这样说还不是很清楚，让我们来看个比较完整的例子吧：

Node* find_and_delete(Node *head, int target)
{
    Node *prev = NULL;
    Node *entry = head;
    while (entry != NULL) {
        if (entry->val == target) {
            if (prev) {
                prev->next = entry->next;
            } else {
                head = entry->next;
            }   
            break;
        }   
        prev = entry;
        entry = entry->next;
    }   
    return head;
}

函数

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find_and_delete
完成了这么一个功能：遍历一个单向链表，如果找到和给出的target相等的值，就从链表中删除这个节点。上述方法是一般c语言教科书（例如谭浩强之类）里的标准做法。但是这种方法遭到了Linus大神的强烈鄙视和唾弃，认为这是不懂指针的人的做法。Linus提到了一个使用二级指针的方法
1

*pp = entry->next
可以让我们不需要判断就可以删除，这是怎么做到的呢？

void find_and_delete2(Node **head, int target)
{
    while (*head != NULL) {
        Node *entry = *head;
        if (entry->val == target) {
            *head = entry->next;
            break;
        }   
        head = &(entry->next);
    }   
}

在

1

find_and_delete2
函数中，巧妙的使用了一个二级指针，从而直接修改了当前entry指针的指向，代码精简了很多。不过这个代码并非很容易正确实现（大家可以自己试试，能不能一遍写正确）。(特别的，想想
1

head = &(entry->next)，
和
1

*head = entry->next
有什么区别？ )

不得不说玩kernel的大神对指针的理解确实比我辈强的多，学一招还挺有用的。学了这招以后去leetcode上切了一题，也用了二级指针:-)

https://github.com/codescv/leetcode/blob/master/AddTwoNumbersNoNew.cpp

最后，上个完整的测试代码例子，供懒得敲代码的人玩耍：

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

struct NodeStruct {
    int val;
    struct NodeStruct *next;
};

typedef struct NodeStruct Node;

Node* find_and_delete(Node *head, int target)
{
    Node *prev = NULL;
    Node *entry = head;
    while (entry != NULL) {
        if (entry->val == target) {
            if (prev) {
                prev->next = entry->next;
            } else {
                head = entry->next;
            }
            break;
        }
        prev = entry;
        entry = entry->next;
    }
    return head;
}

void find_and_delete2(Node **head, int target)
{
    while (*head != NULL) {
        Node *entry = *head;
        if (entry->val == target) {
            *head = entry->next;
            break;
        }
        head = &(entry->next);
    }
}

Node *make_list(int a[], int n)
{
    Node *result = NULL;
    Node *runner;
    int i;
    if (n == 0) {
        return NULL;
    }
    result = (Node *)malloc(sizeof(Node));
    result->val = a[0];
    runner = result;
    for (i = 1; i < n; i++) {
        runner->next = (Node *)malloc(sizeof(Node));
        runner->next->val = a[i];
        runner = runner->next;
    }
    return result;
}

void print_list(Node *l)
{
    while (l) {
        printf("%d", l->val);
        if (l->next) {
            printf("->");
        } else {
            printf("\n");
        }
        l = l->next;
    }
}

int main(int argc, const char *argv[])
{
    int a[] = {1,2,3,4,5};
    Node *l = make_list(a, 5);
    print_list(l);
    l = find_and_delete(l, 3);
    print_list(l);
    l = find_and_delete(l, 1);
    print_list(l);
    Node *l2 = make_list(a,5);
    print_list(l2);
    find_and_delete2(&l2, 3);
    print_list(l2);
    find_and_delete2(&l2, 1);
    print_list(l2);
    return 0;
}