c++11 - clang++失败，但g++成功在赋值中使用强制转换为const-unrelated-type运算符

Question

这是一个简短的示例，再现了“no viable conversion” with lemon for clang but valid for g++在编译器行为上的差异。

#include <iostream>

struct A { 
    int i; 
};

#ifndef UNSCREW_CLANG
using cast_type = const A;
#else 
using cast_type = A;
#endif

struct B {
    operator cast_type () const {
        return A{i};
    }
    int i;
}; 

int main () { 
    A a{0};
    B b{1};

#ifndef CLANG_WORKAROUND
    a = b;
#else    
    a = b.operator cast_type ();
#endif    

    std::cout << a.i << std::endl;    

    return EXIT_SUCCESS;
}

在bolt的上 live

g++(4.9，5.2)会以静默方式进行编译；而clang++(3.5，3.7)编译它

如果

using cast_type = A;

或者

using cast_type = const A;
// [...] 
a = b.operator cast_type ();

被使用，
，但默认值不是

using cast_type = const A;
// [...] 
a = b; 

在这种情况下，clang++(3.5)会怪 a = b:

testling.c++:25:9: error: no viable conversion from 'B' to 'A'
    a = b;
        ^
testling.c++:3:8: note: candidate constructor (the implicit copy constructor) 
not viable:
      no known conversion from 'B' to 'const A &' for 1st argument
struct A { 
       ^
testling.c++:3:8: note: candidate constructor (the implicit move constructor) 
not viable:
      no known conversion from 'B' to 'A &&' for 1st argument
struct A { 
       ^
testling.c++:14:5: note: candidate function
    operator cast_type () const {
    ^
testling.c++:3:8: note: passing argument to parameter here
struct A { 

关于2011¹标准:clang++是否可以拒绝默认代码，或者g++是否可以接受默认代码？

注意::这是 ，而不是，是关于 const上的 cast_type限定词是否有意义的问题。这是关于哪个编译器工作符合标准的 ，并且仅关于那个。

¹2014年应该不会有所作为。

编辑:

请避免使用通用c++标记重新标记。
我首先想知道哪种行为符合2011年标准，并且暂时将委员会的奉献 not to break existing (< 2011) code保留在ansatz之内。

Answer 1

因此，此clang错误报告rvalue overload hides the const lvalue one? 似乎涵盖了此示例，该示例具有以下示例:

struct A{};
struct B{operator const A()const;};
void f(A const&);
#ifdef ERR
void f(A&&);
#endif
int main(){
  B a;
  f(a);
}

失败，并显示与OP的代码相同的错误。理查德·史密斯(Richard Smith)最后说:

Update: we're correct to choose 'f(A&&)', but we're wrong to reject the initialization of the parameter. Further reduced:

  struct A {};
  struct B { operator const A(); } b;
  A &&a = b;

Here, [dcl.init.ref]p5 bullet 2 bullet 1 bullet 2 does not apply, because [over.match.ref]p1 finds no candidate conversion functions, because "A" is not reference-compatible with "const A". So we fall into [dcl.init.ref]p5 bullet 2 bullet 2, and copy-initialize a temporary of type A from 'b', and bind the reference to that. I'm not sure where in that process we go wrong.

但由于 defect report 1604之后又返回了另一条评论:

DR1604 changed the rules so that

 A &&a = b;

is now ill-formed. So we're now correct to reject the initialization. But this is still a terrible answer; I've prodded CWG again. We should probably discard f(A&&) during overload resolution.

因此，似乎clang在今天基于标准语言在技术上做对了，但是它可能会发生变化，因为至少从clang团队看来这是正确的结果。因此，大概这将导致缺陷报告，我们必须等到问题解决后才能得出最终结论。

更新

好像 defect report 2077是基于此问题提交的。