模板偏特擴展到外類型的結果在歧義

Question

我試圖編譯克++下面的程序 - 4.7（20120228-1）：

#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; 
} 

這導致歧義如下：

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>' 

這是因爲一種奇怪的可變參數參數擴建工程的論證組的簡單擴大，而是作爲一個可變參數包擴展到嵌套一些其他的模板類型中一旦出現故障。

這是簡單的編譯器瘋狂還是我做什麼可怕的錯誤？

Answer 1

它看起來像一個GCC的bug，因爲不確定性消失，如果你有一個固定數量的參數替換參數包：http://ideone.com/6D4Fi。

這類含糊不清的也可以使用std::enable_if解決。

/* 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 >