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.
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