当我取消注释下面的main()代码行时,Visual Studio 2015无法编译(在其他情况下可以编译代码)。
#include <iostream>
#include <type_traits>
template <typename Output, std::size_t... Input> struct RemoveLastHelper;
template <template <std::size_t...> class Z, std::size_t... Accumulated, std::size_t First, std::size_t... Rest>
struct RemoveLastHelper<Z<Accumulated...>, First, Rest...> :
RemoveLastHelper<Z<Accumulated..., First>, Rest...> {};
template <template <std::size_t...> class Z, std::size_t... Accumulated, std::size_t Last>
struct RemoveLastHelper<Z<Accumulated...>, Last> {
using type = Z<Accumulated...>;
};
template <template <std::size_t...> class Z, std::size_t... Is>
using RemoveLast = RemoveLastHelper<Z<>, Is...>;
struct Base {
virtual ~Base() = default;
virtual void foo() const = 0;
};
template <std::size_t... Is> struct Derived;
template <std::size_t R>
struct Derived<R> : public Base {
virtual void foo() const override {}
};
// For example, Derived<R,D1,D2,D3> inherits from Derived<R,D1,D2>, which inherits from
// Derived<R,D1>, which inherits from Derived<R>, which inherits from Base (by the above).
template <std::size_t R, std::size_t... Is>
struct Derived<R, Is...> : public RemoveLast<Derived, R, Is...>::type {
virtual void foo() const override {RemoveLast<Derived, R, Is...>::type::foo();}
};
int main() {
// Derived<0,2,1> r; // This line won't compile.
}
它出现了错误:
'Derived<0,2>': invalid template argument for template parameter 'Z', expected a class template
'RemoveLast': left of '::' must be a class/struct/union
'type': is not a class or namespace name
那么问题出在哪里呢?GCC 5.1可以编译它,但我不知道那是否只是偶然。这个语法
virtual void foo() const override {RemoveLast<Derived, R, Is...>::type::foo();}
合法吗?如果不合法,我应该如何实现它(我只是想调用它的基类foo()函数)。如果它是合法的,我应该如何重写它,以便Visual Studio可以接受它(我需要它能在Visual Studio上工作)?