模板继承可以从它的另一个特化是一个循环依赖,还是一个clang的bug

Question

考虑以下代码:

#include <iostream>
#include <utility>
#include <tuple>

template <class TargetIndices, class SourceIndices>
struct assign;

template <size_t... TargetIs, size_t... SourceIs>
struct assign<std::index_sequence<TargetIs...>, std::index_sequence<SourceIs...>>: assign<std::index_sequence<TargetIs>, std::index_sequence<SourceIs>>... {
   template <class TargetTuple, class SourceTuple>
   assign(TargetTuple &target, const SourceTuple &source): assign<std::index_sequence<TargetIs>, std::index_sequence<SourceIs>>(target, source)... { }
};

template <size_t FirstTI, size_t FirstSI>
struct assign<std::index_sequence<FirstTI>, std::index_sequence<FirstSI>> {
   template <class TargetTuple, class SourceTuple>
   assign(TargetTuple &target, const SourceTuple &source) {
      std::get<FirstTI>(target) = std::get<FirstSI>(source);
   }
};


int main() {
   std::tuple<int, int, int> t1 = std::make_tuple(0,0,0), t2 = std::make_tuple(1,2,3);
   assign<std::index_sequence<0,2>, std::index_sequence<2,0>>(t1, t2);
   std::cout << std::get<0>(t1) << " " << std::get<1>(t1) << " " << std::get<2>(t1) << std::endl;
}

clang ++声明继承中存在循环依赖.g ++编译它没有任何问题.哪个编译器是对的？

Answer 1

它看起来像一个铿锵的bug.

最小的复制看起来像这样:

template <int...> struct q {};

template <int... Is> using w = q<Is...>;

template <class T>
struct a;

template <int F>
struct a<w<F>> {};

template <int... T>
struct a<w<T...>>: a<w<T>>... {}; // g++ OK, clang++ breaks

模板别名在这里是必不可少的.在最后一行使用原始模板而不是模板别名时,它会编译:

struct a<q<T...>>: a<q<T>>... {}; // g++, clang++ OK