13

这是另一个问题的后续。它指的是同一个问题（我希望），但使用了一个完全不同的例子来说明它。原因是在前面的示例中，只有实验性 GCC 4.9 因编译器错误而失败。在此示例中，Clang 和 GCC 4.8.1 也以不同的方式失败：Clang 产生了意外结果，而 GCC 4.8.1 报告了不同的错误消息。

上一个问题的答案或多或少说代码是有效的，问题出在 GCC 的实验版本上。但这个结果让我更加怀疑。几个月来，我一直被我怀疑相关（或相同）的问题所困扰，这是我第一次有一个小的具体例子来说明。

所以，这里有一些代码。首先，一些通用代码将 SFINAE 应用于由可变参数模板别名元函数指定的任意测试F：

#include <iostream>
using namespace std;

using _true  = integral_constant <bool, true>;
using _false = integral_constant <bool, false>;

template <typename T> using pass = _true;

template <template <typename...> class F>
struct test
{
    template <typename... A> static _false           _(...);
    template <typename... A> static pass <F <A...> > _(int);
};

template <template <typename...> class F, typename... A>
using sfinae = decltype(test <F>::template _<A...>(0));

其次，一个特定的测试，检查给定的类是否定义了一个名为的类型type：

template <typename T> using type_of  = typename T::type;
template <typename T> using has_type = sfinae <type_of, T>;

最后，举个例子：

struct A { using type = double; };

int main()
{
    cout << has_type <int>() << ", ";
    cout << has_type <A>()   << endl;
}

预期的结果是0, 1。叮当说0, 0。GCC 4.8.1 说

tst.cpp: In substitution of ‘template<class T> using type_of = typename T::type [with T = A ...]’:
tst.cpp:15:51: required from ‘struct test<type_of>’
tst.cpp:19:67: required by substitution of ‘template<template<class ...> class F, class ... A> using sfinae = decltype (test:: _<A ...>(0)) [with F = type_of; A = {T}]’
tst.cpp:24:58: required from here
tst.cpp:23:56: error: ‘A ...’ is not a class, struct, or union type
  template <typename T> using type_of = typename T::type; 
                                                        ^

和 GCC 4.9 说

tst.cpp:19:67:   required by substitution of ‘template<template<class ...> class F, class ... A> using sfinae = decltype (test:: _<A ...>(0)) [with F = type_of; A = {T}]’
tst.cpp:24:58:   required from here
tst.cpp:15:51: error: pack expansion argument for non-pack parameter ‘T’ of alias template ‘template<class T> using type_of = typename T::type’
  template <typename... A> static pass <F <A...> > _(int);
                                                   ^

（行号可能会有所不同）。所以，一切都失败了，以不同的方式。

现在，这里有一个解决方法。元函数car从给定包中选择第一种类型，然后将测试重新定义为type_of2，现在是可变参数：

template <typename... T> struct car_t;
template <typename... T> using  car = type_of <car_t <T...> >;

template <typename T, typename... Tn>
struct car_t <T, Tn...> { using type = T; };

template <typename... T> using type_of2  = typename car <T...>::type;
template <typename T>    using has_type2 = sfinae <type_of2, T>;

int main()
{
    cout << has_type2 <int>() << ", ";
    cout << has_type2 <A>()   << endl;
}

现在所有三个编译器都0, 1按预期说。有趣的是，对于任何版本的 GCC，我们都必须删除has_type（即使我们不使用它）并且只留下has_type2; 否则我们有类似的错误。

总结一下：我看到了一个模板的问题，它期望表单的可变参数模板参数

template <typename...> class F

我们实际上将表单的非可变模板别名作为输入

template <typename T> using alias = // ... anything including T or not

最后F像可变参数一样调用：

F <A...>

到目前为止的意见说这是有效的，但现在似乎三个编译器不同意。所以问题又来了：它有效吗？

对我来说这很重要，因为我有几十个现有代码文件，假设这是有效的，现在我无论如何都需要重新设计（因为这些编译器存在实际问题），但确切的重新设计将取决于答案。

4

2 回答 2

2

This does not answer the question whether the code above is valid, but is a quite pretty workaround that I have found by experimenting shortly after asking the question, and I think is useful to share.

All that is needed are the following definitions:

template <template <typename...> class F>
struct temp { };

template <typename... A, template <typename...> class F>
F <A...> subs_fun(temp <F>);

template <template <typename...> class F, typename... A>
using subs = decltype(subs_fun <A...>(temp <F>()));

then, wherever F <A...> would be problematic, replace it with subs <F, A...>. That's it. I cannot explain why, but it has worked in all cases so far.

For instance, in the SFINAE example of the question, just replace line

template <typename... A> static pass <F <A...> > _(int);

by

template <typename... A> static pass <subs <F, A...> > _(int);

This is a change at one point only, all remaining code stays the same. You don't need to redefine or wrap every template metafunction that with be used as F. Here's a live example.

If F <A...> is indeed valid and compilers support it eventually, it is again easy to switch back because changes are minimal.

I find this important because it allows specifying a SFINAE test in just two lines

template <typename T> using type_of  = typename T::type;
template <typename T> using has_type = sfinae <type_of, T>;

and is completely generic. Typically, each such test needs at least 10 lines of code and implementations of <type_traits> are full of such code. In some cases such code blocks are defined as macros. With this solution, templates can do the job and macros are not needed.

于 2014-03-13T09:35:55.260 回答
0

我认为情况非常标准化。C++11 14.3.3/1 说：

模板模板参数的模板参数应该是类模板或别名模板的名称，表示为id-expression。

于 2013-11-29T18:19:21.437 回答