返回值和局部变量的区别

Question

假设我有

#include <string>

class A
{
public:
    template<class T>
    operator T();

    A child();
};

void f()
{
    A a;
    std::string s1 = a;            // ok
    std::string s2 = a.child();    // error (line 34)

    s1 = a;           // error (line 36)
    s2 = a.child();   // error (line 37)

}

std::string 构造函数可以采用 char* 或 std::string 引用，这就是赋值不明确的原因。但是为什么我的编译器（VC++10）抱怨第二个赋值而不是第一个呢？

我正在寻找一种方法来优先考虑模板转换运算符而不是重载的构造函数。

我收到以下错误：

1>------ Build started: Project: Plasma4Test, Configuration: Debug Win32 ------
1>  Plasma4Test.cpp
1>d:\bitbucket\vx\projects\plasma4test\plasma4test.cpp(34): error C2440: 'initializing' : cannot convert from 'A' to 'std::basic_string<_Elem,_Traits,_Ax>'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          No constructor could take the source type, or constructor overload resolution was ambiguous
1>d:\bitbucket\vx\projects\plasma4test\plasma4test.cpp(36): error C2593: 'operator =' is ambiguous
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(772): could be 'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(_Elem)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(767): or       'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(const _Elem *)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(762): or       'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(const std::basic_string<_Elem,_Traits,_Ax> &)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          while trying to match the argument list '(std::string, A)'
1>d:\bitbucket\vx\projects\plasma4test\plasma4test.cpp(37): error C2593: 'operator =' is ambiguous
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(772): could be 'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(_Elem)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(767): or       'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(const _Elem *)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(762): or       'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(const std::basic_string<_Elem,_Traits,_Ax> &)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          c:\program files\microsoft visual studio 10.0\vc\include\xstring(707): or       'std::basic_string<_Elem,_Traits,_Ax> &std::basic_string<_Elem,_Traits,_Ax>::operator =(std::basic_string<_Elem,_Traits,_Ax> &&)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Ax=std::allocator<char>
1>          ]
1>          while trying to match the argument list '(std::string, A)'
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

Answer 1

对我来说这似乎是一个 VC10 错误，它与std::string不相关。

BUG 隔离：

我把它归结为下面的例子：

#include <string>

class B
{
public:
    B(char const*) { }
    B(B&&) { }
};

class A
{
public:
    operator char* const () { return 0; }
    operator B () { return B(0); }
};

int main()
{
    A a;
    B b1 = a; // fine
    B b2 = A(); // error C2440: 'initializing' : cannot convert from 'A' to 'B'
                // No constructor could take the source type, or constructor
                // overload resolution was ambiguous.
}

B 类有一个移动构造函数和一个采用const char* 的构造函数。当尝试从rvalue 初始化b2 时，VC10 似乎无法选择转换运算符为B。

Clang 3.2 和 GCC 4.7.2 都选择转换运算符为B。

C++ 标准规则：

C++ 标准第 8.5/16 条规定：

[对于复制初始化的这种情况，]“用户定义的转换序列，可以从源类型转换为目标类型或（当使用转换函数时）到其派生类如 13.3.1.4 所述枚举，并通过重载决议选择最佳的一个（13.3）"

如果我们考虑我们示例中从源类型 (A) 到目标类型 (B) 的所有可用转换序列，那么涉及到 A 的用户定义转换函数到 @ 987654332@ 需要进一步 转换（通过接受char const* 的B 的构造函数完成）才能到达目标类型 B。因此，它比使用A 的用户定义转换函数到B（根据13.3.3.2）的那一步要长一步，这使得后者更可取。

这似乎证实这是一个 VC10 错误。