为什么允许指向不完整类型的指针而不是不完整类型的变量? [英] Why are pointers to incomplete types allowed and not variables of incomplete types?

问题描述

为什么以下是合法的:

typedef struct a aType;
struct a
{
   int x;
   aType *b;
};

以及以下违法行为:

void main()
{
    typedef struct a aType;
    aType someVariable;
    struct a
    {
      int x;
      aType *b;
    };
}

我只是好奇，因为在每种情况下它都是前向引用，据我所知，至少对于函数和变量，前向引用是不合法的.

I'm just curious as in each case it is forward referencing and as far as I know, at least for functions and variables, forward referencing is not legal.

另外，C++ 的答案是否也一样?

Also, would the answer for this be the same for C++ as well?

推荐答案

就这样:

typedef struct a aType;
struct a { int x; aType *b; };

等同于:

struct a;
typedef struct a aType;
struct a { int x; aType *b; };

所以你正向声明一个struct，typedef然后定义它.完全没问题.

So you're forward-declaring a struct, typedefing it and later defining it. Perfectly fine.

现在是第二个例子:

typedef struct a aType;
aType someVariable;
struct a { int x; aType *b; };

这与它在本地范围内的事实无关.这是怎么回事:

This has nothing to do with the fact that it's in local scope. What's happening is this:

struct a;
typedef struct a aType;
aType someVariable; // error: when it gets here, aType is still incomplete
struct a { int x; aType *b; };
aType someVariable; // perfectly fine, aType not incomplete

记住编译是按顺序进行的.当你尝试声明 someVariable 时，编译器还不知道 struct a 是什么，所以它不知道它的大小，因此它不知道有多少内存为它分配，因此编译错误.在定义 aType 之后声明它按预期工作.

Remember that compilation happens in order. When you try to declare someVariable the compiler doesn't know what struct a is yet, so it doesn't know its size, hence it doesn't know how much memory to allocate for it, hence a compile error. Declaring it after aType is defined works as expected.

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