为什么不能一个结构是自身的成员？ [英] Why can't a struct be a member of itself?

问题描述

我有一个结构富。声明类型富* 的成员如下：

  typedef结构美孚
{
    结构* foo的孩子[26];
} foo的;
 

但是，如果我试图声明类型的成员富我得到一个错误：

  typedef结构美孚
{
    结构foo的孩子[26];
} foo的;
 

本声明给我的错误

  

结构富'的定义是不完整的，直到结束的'}

解决方案

一个结构 T 不能包含自身。你怎么知道它的大小？这将是不可能这样做的，因为 T 的大小将要求您知道 T 的大小（因为 T 包含另一个 T ）。这变成一个无限递归。

您可以有一个指针 T 在结构内部 T 因为一个指针的大小是不一样的尺寸为指向的对象：在这种情况下，你只储存的内存地址在另一个 T 存储 - 所有你需要做的空间基本上是空间，你需要存储的内存地址在另一个 T 生活。

I have a struct foo. Declaring a member of type foo* works:

typedef struct foo
{
    struct foo* children[26];
} foo;

But if I try to declare a member of type foo I get an error:

typedef struct foo
{
    struct foo children[26];
} foo;

This declaration gives me the error

definition of 'struct foo' is not complete until the closing '}'

解决方案

A structure T cannot contain itself. How would you know its size? It would be impossible to do so, because the size of T would require you to know the size of T (because T contains another T). This turns into an infinite recursion.

You can have a pointer to T inside a structure T because the size of a pointer is not the same size as the pointed-to object: in this case, you would just store an address of memory where another T is stored - all the space you need to do that is basically the space you need to store a memory address where another T lives.

这篇关于为什么不能一个结构是自身的成员？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！