问题描述
更新
我发布了中可能有一个 static_assert ,这完全破坏了我的元功能。
Basically I want to check if a templated type T<U, Args...> can be constructed from some other type T<V, Args...>. Where T and Args... is the same in both types. The problem is, T<> might have a static_assert in it that totally breaks my metafunction.
下面是我要做的事情的简要摘要。
Below is a rough summary of what I'm trying to do.
template<typename T> struct fake_alloc { using value_type = T; }; template<typename T, typename Alloc = fake_alloc<T>> struct fake_cont { using value_type = T; // comment the line below out, and it compiles, how can I get it to compile without commenting this out??? static_assert(std::is_same<value_type, typename Alloc::value_type>::value, "must be the same type"); }; template<typename T, typename U, typename = void> struct sample_rebind { using type = T; }; template<template<typename...> class Container, typename T, typename U, typename... OtherArgs> struct sample_rebind< Container<T, OtherArgs...>, U, std::enable_if_t< std::is_constructible< Container<T, OtherArgs...>, Container<U, OtherArgs...> >::value > > { using type = Container<U, OtherArgs...>; }; static_assert( std::is_same< fake_cont<int, fake_alloc<int>>, typename sample_rebind<fake_cont<int>, double>::type >::value, "This should pass!" );
您可以看到所需的行为是最终的 static_assert 应该会通过,但是不幸的是,它甚至还没有达到那个程度,因为 fake_cont 中的 static_assert 当 std :: is_constructible<> 尝试调用 fake_cont 的构造函数时触发。
As you can see the desired behavior is that the final static_assert should pass, but unfortunately, it doesn't even get to that point as the static_assert in fake_cont is triggered when std::is_constructible<> attempts to call fake_cont's constructor.
在真实代码 fake_cont 中是libc ++的 std :: vector ,所以我可以't修改它的胆量,或 std :: is_constructible 的胆量。
In the real code fake_cont is libc++'s std::vector, so I can't modify it's guts, or std::is_constructible's guts.
任何解决此特定问题的建议对此问题表示赞赏,特别建议您对SFINAE的一般建议在 static_assert 左右。
Any advice for working around this specific issue is appreciated, and any advice in general for SFINAE'ing around static_assert's is especially appreciated.
编辑:is_same的第一部分应该是 fake_cont< int,fake_alloc< int>>
the first part of the is_same should have been fake_cont<int, fake_alloc<int>>
编辑2 :如果您注释掉 fake_cont 中的 static_assert ,它将编译(clang 3.5)。这就是我想要的。所以我只需要一些方法来避免 fake_cont 中的 static_assert 。
Edit 2: If you comment out the static_assert in fake_cont, it compiles (clang 3.5). And that's what I want. So I just need some way to avoid the static_assert in fake_cont.
推荐答案
namespace details { template<class T,class=void> struct extra_test_t: std::true_type {}; }
然后我们对以下项进行额外的测试:
We then fold an extra test in:
template<class...>struct types{using type=types;}; template<template<typename...> class Container, typename T, typename U, typename... OtherArgs> struct sample_rebind< Container<T, OtherArgs...>, U, std::enable_if_t< details::extra_test_t< types< Container<T, OtherArgs...>, U > >::value && std::is_constructible< Container<T, OtherArgs...>, Container<U, OtherArgs...> >::value > > { using type = Container<U, OtherArgs...>; };
,然后编写额外的测试:
and we write the extra test:
namespace details { template<class T, class Alloc, class U> struct extra_test_t< types<std::vector<T,Alloc>, U>, typename std::enable_if< std::is_same<value_type, typename Alloc::value_type>::value >::type > : std::true_type {}; template<class T, class Alloc, class U> struct extra_test_t< types<std::vector<T,Alloc>, U>, typename std::enable_if< !std::is_same<value_type, typename Alloc::value_type>::value >::type > : std::false_type {}; }
基本上,这使我们可以在测试中注入补丁以匹配 static_assert 。
basically, this lets us inject "patches" on our test to match the static_assert.
如果我们有 is_std_container< T> 和 get_allocator< T> ,我们可以这样写:
If we had is_std_container<T> and get_allocator<T>, we could write:
namespace details { template<template<class...>class Z,class T, class...Other, class U> struct extra_test_t< types<Z<T,Other...>>, U>, typename std::enable_if< is_std_container<Z<T,Other...>>>::value && std::is_same< value_type, typename get_allocator<Z<T,Other...>>::value_type >::value >::type > : std::true_type {}; template<class T, class Alloc, class U> struct extra_test_t< types<std::vector<T,Alloc>, U>, typename std::enable_if< is_std_container<Z<T,Other...>>>::value && !std::is_same< value_type, typename get_allocator<Z<T,Other...>>::value_type >::value >::type > : std::false_type {}; }
或者我们可以声明带有 allocator_type的任何内容可能无法反弹。
or we could just state that anything with an allocator_type probably cannot be rebound.
对这个问题更了解容器的方法是提取分配器类型( ::: allocator_type ),并将容器参数列表中所有分配器类型的实例替换为 T 与的重新绑定U 。这仍然很棘手,因为 std :: map< int,int> 具有类型为 std :: allocator<的分配器。 std :: pair< const int,int> > ,并且不能区分键 int 和值 int
A more container-aware approach to this problem would be to extract the allocator type (::allocator_type), and replace all instances of the allocator type in the container argument list with a rebind of T to U somehow. This is still tricky, as std::map<int, int> has an allocator of type std::allocator< std::pair<const int, int> >, and distinguishing between the key int and the value int isn't possible in a generic way.
这篇关于如何使SFINAE发挥static_assert的作用的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!