c中的空指针是什么 [英] what is a null pointer in c

问题描述

#include int main(void){结构体 findEntry{整数值；结构体入口 *next;};结构体条目 n1、n2、n3；struct entry *list_pointer = &n1;n1.value = 100;n1.next = &n2;n2.value = 200;n2.next = &n2;n3.value = 300;n3.next = (stuct entry *) 0;while (list_pointer != (struct entry *)0) {printf("%i\n", list_pointer->value);list_pointer = list_pointer->next;}返回0；}

我不明白语法 (struct entry *) 0 在这里是什么意思.在我正在阅读的书中，它说它是一个空指针.我尝试在网上查找，但我不知道该输入什么.我得到的NULL 指针"的 Google 搜索结果与我预期的不同.

解决方案

任何值为 0 的指针类型都称为空指针.以下是 C 标准 §6.3.2.3 的解释:

<块引用>

3. 值为 0 的整数常量表达式，或这样的表达式强制转换为 void * 类型，称为空指针常量.如果一个空指针常量被转换为指针类型，结果指针，称为空指针，保证比较不等于指向任何对象或函数的指针

#include <stdio.h>

int main(void)
{
    struct findEntry
    {
        int value;
        struct entry *next;
    };

    struct entry n1, n2, n3;
    struct entry *list_pointer = &n1;

    n1.value = 100;
    n1.next = &n2;

    n2.value = 200;
    n2.next = &n2;

    n3.value = 300;
    n3.next = (stuct entry *) 0;

    while (list_pointer != (struct entry *)0) {
        printf("%i\n", list_pointer->value);
        list_pointer = list_pointer->next;
    }
    return 0;
}

I don't understand what the syntax (struct entry *) 0, means here. In the book I am reading it says it is a null pointer. I tried looking online but I didn't know exactly what to type. The Google search results I got for 'NULL pointer' were different from what I expected.

解决方案

Any pointer type with the value 0 is called a null pointer. Here is the explanation from the C standard §6.3.2.3:

3. An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant. If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function

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