如何声明相互引用的类?
自从我使用 C++ 已经很长时间了，我在类相互引用时遇到了一些麻烦.
It's been a long time since I've done C++ and I'm running into some trouble with classes referencing each other.
现在我有类似的东西:
a.h
class a
{
public:
a();
bool skeletonfunc(b temp);
};
b.h
class b
{
public:
b();
bool skeletonfunc(a temp);
};
由于每个人都需要对另一个人的引用，我发现我不能在顶部执行彼此的#include，否则我最终会陷入各种奇怪的循环.
Since each one needs a reference to the other, I've found I can't do a #include of each other at the top or I end up in a weird loop of sorts with the includes.
那么我怎样才能让 a 可以使用 b 反之亦然，而不会造成循环 #include 问题?
So how can I make it so that a can use b and vice versa without making a cyclical #include problem?
谢谢！
推荐答案
你必须使用前向声明:
a.h
class b;
class a
{
public:
a();
bool skeletonfunc(b temp);
}
然而，在许多情况下，这会迫使您在方法调用或成员变量中使用引用或指针，因为您不能在两个类头中都拥有完整的类型.如果必须知道类型的大小，则需要使用引用或指针.但是，如果只需要方法声明，则可以使用该类型.
However, in many situations, this can force you to work with references or pointers in your method calls or member variables, since you can't have the full types in both class headers. If the size of the type must be known, you need to use a reference or pointer. You can, however, use the type if only a method declaration is required.
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