如何实现两个可以互相访问的结构?

Say*_*iss 8 c++ syntax dependencies struct forward-declaration

我写的代码:

    struct A;
    struct B;
    struct A
    {
        int v;
        int f(B b)
        {
            return b.v;
        }
    };

    struct B
    {
        int v;
        int f(A a)
        {
            return a.v;
        }
    };
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编译消息:

|In member function 'int A::f(B)':|
11|error: 'b' has incomplete type|
7|error: forward declaration of 'struct B'|
||=== Build finished: 2 errors, 0 warnings (0 minutes, 0 seconds) ===|
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我知道,为什么代码不正确,但我不知道如何实现两个可以相互访问的结构.有什么优雅的方式吗?提前致谢.

Tem*_*Rex 16

如果要保留成员函数的完全相同的签名,则必须推迟成员函数的定义,直到看到两个类定义为止

    // forward declarations
    struct A;
    struct B;

    struct A
    {
        int v;
        int f(B b); // works thanks to forward declaration
    };

    struct B
    {
        int v;
        int f(A a);
    };

    int A::f(B b) { return b.v; } // full class definition of B has been seen
    int B::f(A a) { return a.v; } // full class definition of A has been seen
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您也可以使用const&函数参数(更好的表现时间A和更大B),但即使这样,您也必须推迟函数定义,直到看到两个类定义.