自指结构声明 [英] Self referring structure declaration

问题描述

该follwing声明是有效的。

 结构节点
{
    int类型的;
    结构节点*接下来的;
};
 

然而，当我们定义了下面，它给错误。

 错误：场下一步具有不完整的类型
 

为什么会这样呢？

 结构节点
{
    int类型的;
    结构节点旁边; / *不是指针* /
};
 

解决方案

在结构节点节点是一个结构标签，这在点你写它创建了一个不完全类型：一个结构变量，它是不是在宣告这一点，但没有定义。该类型是不完整之前的最后}; 你的结构的

在C，一个不完全类型可即使是在它完全定义，使用指针该类型引用。然而，你不能分配类型的变量（实例），因为实际的结构定义尚未确定。 （它的工作原理完全一样的C ++抽象基类，如果你熟悉的。）

所以，当你写

 结构节点{
  int类型的;
  结构节点*接下来的;
};
 

行结构节点*下一表示这里是一个指向一个结构节点，尽管我不知道该类型是如何定义的呢。但是你不能声明类型的变量结构节点这非常相同类型的结构定义里面，仅仅是因为你不能使用的东西你已经创建它。

The follwing declaration is valid.

struct node
{
    int a;
    struct node *next;
};

However, when we define the following, it gives error.

"error: field ‘next’ has incomplete type"

Why is it so?

struct node
{
    int a;
    struct node next; /* Not a pointer */   
};

解决方案

node in struct node is a "struct tag", which at the point you write it creates an "incomplete type": a struct variable which is not at this point declared, but not defined. The type is not complete before the final }; of your struct.

In C, an incomplete type can be referenced even before it is fully defined, by using a pointer to that type. You can however not allocate a variable (instance) of that type, because the actual struct definition is yet to be defined. (It works exactly like abstract base classes in C++, if you are familiar with those.)

So when you write

struct node {
  int a;
  struct node *next;
};

the row struct node *next means "here is a pointer to a struct node, even though I have no idea how that type is defined yet". But you cannot declare a variable of type struct node inside the struct definition of that very same type, simply because you cannot use something before you have created it.

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