使用别名模板时,无法将"std :: unique_ptr"分配给clang中的基类

Question

以下代码在gcc 4.9.3和clang 3.7.1上编译并运行得很好

// std::unique_ptr
#include <memory>

// Template class for template-template arguments
template <typename Real>
struct Bar {};

// Base class 
template <typename T,template <typename> class XX>
struct Base {};

// Derived class that operates only on Bar 
template <typename Real>
struct Derived : public Base <Real,Bar> {};

// Holds the unique_ptr 
template <typename T,template <typename> class XX>
struct Foo {
    std::unique_ptr <Base <T,XX>> foo;
};

// Create an alias template 
template <typename Real>
using Buz = Bar <Real>;

int main() {
    #if 0
    auto f = Foo <double,Buz> (); //Causes error!
    #else
    auto f = Foo <double,Bar> ();
    #endif
    f.foo =  std::make_unique <Derived <double>> (Derived <double>());
}

但是,如果我们更改为#if 0to #if 1,gcc编译,但clang不会:

g++ -std=c++14 test03.cpp -o test03_gcc
clang++ -std=c++14 test03.cpp -o test03_clang
test03.cpp:32:11: error: no viable overloaded '='
    f.foo =  std::make_unique <Derived <double>> (Derived <double>());
    ~~~~~ ^  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/lib/gcc/x86_64-pc-linux-gnu/4.9.3/include/g++-v4/bits/unique_ptr.h:249:7: note: 
      candidate function not viable: no known conversion from
      'unique_ptr<Derived<double>, default_delete<Derived<double>>>' to
      'unique_ptr<Base<double, Buz>, default_delete<Base<double, Buz>>>' for
      1st argument
      operator=(unique_ptr&& __u) noexcept
      ^
/usr/lib/gcc/x86_64-pc-linux-gnu/4.9.3/include/g++-v4/bits/unique_ptr.h:278:7: note: 
      candidate function not viable: no known conversion from 'typename
      _MakeUniq<Derived<double> >::__single_object' (aka
      'unique_ptr<Derived<double> >') to 'nullptr_t' for 1st argument
      operator=(nullptr_t) noexcept
      ^
/usr/lib/gcc/x86_64-pc-linux-gnu/4.9.3/include/g++-v4/bits/unique_ptr.h:357:19: note: 
      candidate function not viable: no known conversion from
      'unique_ptr<Derived<double>, default_delete<Derived<double>>>' to
      'const unique_ptr<Base<double, Buz>, default_delete<Base<double,
      Buz>>>' for 1st argument
      unique_ptr& operator=(const unique_ptr&) = delete;
                  ^
/usr/lib/gcc/x86_64-pc-linux-gnu/4.9.3/include/g++-v4/bits/unique_ptr.h:264:22: note: 
      candidate template ignored: disabled by 'enable_if' [with _Up =
      Derived<double>, _Ep = std::default_delete<Derived<double> >]
        typename enable_if< __and_<
                            ^
1 error generated.
Makefile:2: recipe for target 'all' failed
make: *** [all] Error 1

在此上下文中使用别名模板有什么问题？或者,如果gcc比它应该更宽松,为什么会这样呢？

Answer 1

这是CWG问题1244:

14.4 [temp.type]第1段中的例子很重要,

template<template<class> class TT> struct X { };
template<class> struct Y { };
template<class T> using Z = Y<T>;
X<Y> y;
X<Z> z;

并说,y并z具有相同的类型.

只有当别名模板Z被认为等同于类模板时才会出现这种情况Y.但是,14.5.7 [temp.alias]仅描述别名模板的特化,而不是别名模板本身的等效.应该指定这样的规则,这可能很棘手,或者应该删除示例.

我们可以将您的示例缩减为:

std::unique_ptr<Base<double, Buz>> f = 
    std::make_unique<Base<double, Bar>>();

当且仅当Buz且Bar被认为是等同的时,这是良好形成的.gcc认为他们是,clang认为他们不是.关于实际答案是什么仍然是一个悬而未决的问题.