c++ - 非常量复制构造函数和返回值的隐式转换

Question

考虑以下 C++ 代码:

struct B { };
struct A
{
        A(int);
        A(A&); // missing const is intentional
        A(B);
        operator B();
};

A f()
{
        // return A(1); // compiles fine
        return 1; // doesn't compile
}

这在 MSVC++ 2010 上编译得很好(事实上，在 MSVC 上，如果我完全删除 B，它甚至可以工作)。它不适用于 GCC 4.6.0:

conv.cpp: In function ‘A f()’:
conv.cpp:13:9: error: no matching function for call to ‘A::A(A)’
conv.cpp:13:9: note: candidates are:
conv.cpp:6:2: note: A::A(B)
conv.cpp:6:2: note:   no known conversion for argument 1 from ‘A’ to ‘B’
conv.cpp:5:2: note: A::A(A&)
conv.cpp:5:2: note:   no known conversion for argument 1 from ‘A’ to ‘A&’
conv.cpp:4:2: note: A::A(int)
conv.cpp:4:2: note:   no known conversion for argument 1 from ‘A’ to ‘int’

令我困惑的是消息 no known conversion for argument 1 from ‘A’ to ‘B’。考虑到 A::operator B() 定义非常明确，这怎么可能是真的？

Answer 1

因为你不能进行不止一次的隐式转换。你将不得不去 A::A(A::A(int)::operator B()) 来使它工作，而对于编译器来说，这是太多的步骤来弄清楚它是拥有。