7

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 ==========
4

1 回答 1

3

对我来说，这似乎是一个 VC10 错误，与std::string.

错误隔离：

我把它归结为以下示例：

#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.
}

类B有一个移动构造函数和一个带有const char*. 尝试从 初始化b2时rvalue，VC10 似乎无法选择转换运算符到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++ 标准的第 8.5/16 段规定：

[对于复制初始化的这种情况，] “可以从源类型转换为目标类型或（当使用转换函数时）转换为其派生类的用户定义转换序列，如 13.3.1.4 中所述，并且通过重载决议（13.3）选择最好的“

如果我们考虑示例中从源类型A（) 以到达目的地类型。因此，它比使用的用户定义转换函数 to (per 13.3.3.2) 的那一步要长一步，这使得后者更可取。BAchar const*Bchar const* BAB

这似乎证实了这是一个 VC10 错误。

于 2013-01-17T13:37:22.690 回答