可变参数类型的部分模板特化和扩展为外部类型的可变参数包导致歧义

Question

我正在尝试使用 g++-4.7 (20120228-1) 编译以下程序：

#include <cstdlib>
#include <tuple>

template<typename X> struct Y {};

template<typename T, size_t Level, size_t TermLevel> struct A;

// (B) dummy for T=tuple<int, Ts...> just to show it works for simple expansions
template<typename ... Ts, size_t Level, size_t TermLevel>
struct A<std::tuple<int, Ts...>, Level, TermLevel>
{
    A<std::tuple<int, Ts...>, Level+1, TermLevel> value;
};

template<typename ... Ts, size_t Level>
struct A<std::tuple<int, Ts...>, Level, Level> {};


// (C) ambiguous partial specialization
template<typename ... Ts, size_t Level, size_t TermLevel>
struct A<std::tuple<Y<Ts>...>, Level, TermLevel>
{
    A<std::tuple<Y<Ts>...>, Level+1, TermLevel> value;
};

template<typename ... Ts, size_t Level>
struct A<std::tuple<Y<Ts>...>, Level, Level> {};


int main(int argc, const char *argv[])
{
    A<std::tuple<int, float, int>, 0, 5> tint;
    A<std::tuple<Y<int>, Y<float>>, 0, 1> tn;
    return 0;
}

导致歧义如下：

g++-4.7 -g -O0 -std=c++0x    specialization_orig.cc   -o specialization_orig
specialization_orig.cc: In instantiation of 'struct A<std::tuple<Y<int>, Y<float> >, 0ul, 1ul>':
specialization_orig.cc:33:43:   required from here
specialization_orig.cc:23:49: error: ambiguous class template instantiation for 'struct A<std::tuple<Y<int>, Y<float> >, 1ul, 1ul>'
specialization_orig.cc:21:8: error: candidates are: struct A<std::tuple<Y<Ts>...>, Level, TermLevel>
specialization_orig.cc:27:8: error:                 struct A<std::tuple<Y<Ts>...>, Level, Level>
specialization_orig.cc:23:49: error: 'A<std::tuple<Y<Ts>...>, Level, TermLevel>::value' has incomplete type
specialization_orig.cc:6:61: error: declaration of 'struct A<std::tuple<Y<int>, Y<float> >, 1ul, 1ul>'

这有点奇怪，因为可变参数扩展适用于参数包的简单扩展，但一旦可变参数包扩展为嵌套在其他模板类型中，就会失败。

这仅仅是编译器的疯狂还是我做错了什么？

Answer 1

它看起来像一个 GCC 错误，因为如果将参数包替换为固定数量的参数，歧义就会消失：http://ideone.com/6D4Fi。

这种歧义也可以使用std::enable_if来解决。

/* add "typename = void" anonymous parameter which enables individual partial
   specializations by matching the void result in std::enable_if<>::type. */
template<typename T, size_t Level, size_t TermLevel, typename = void> struct A;

// No enable_if needed here: always enabled if the levels are equal.
template<typename ... Ts, size_t Level>
struct A<std::tuple<int, Ts...>, Level, Level, void> {}; // just pass void

// Use enable_if to disable in case the levels are equal.
template<typename ... Ts, size_t Level, size_t TermLevel>
struct A<std::tuple<Y<Ts>...>, Level, TermLevel,
    typename std::enable_if< Level != TermLevel >::type >