可以使用参数包编译c ++ 17 lambda继承的编译器 [英] Compiler which can compile c++17 lambda inheritance with parameter pack

问题描述

我了解了使用函数声明，我想编译最后一个例子。这是：

I read about the using function declaration and I wanted to compile the last example. This is :

#include <iostream>
template <typename... Ts>
struct Overloader : Ts... {
    using Ts::operator()...; // exposes operator() from every base
};

template <typename... T>
constexpr auto make_overloader(T&&... t) {
    return Overloader<T...>{std::forward<T>(t)...};
}

int main() {
    auto o = make_overloader([] (auto const& a) {std::cout << a;},
                             [] (float f) {std::cout << 13 << f;});
}

即使我已经知道并理解它会做什么，我也想编译并测试它。
但是，目前clang-4.0和g ++-7.0似乎都无法对其进行编译。我可以尝试使用任何编译器吗？

Even if I already know and understand what it will do, I would like to compile and test it. However, neither clang-4.0 and g++-7.0 seems to be able to compile it at the moment. Is there any place with any compiler I could try it ?

推荐答案

P0195 ，该语言扩展名允许：

P0195, the proposed language extension that allows for:

template <typename... Ts>
struct Overloader : Ts... {
    using Ts::operator()...; // <== ill-formed without p0195
};

几周前（2016年11月）才在Issaquah被C ++接受。 gcc或clang尚未实现它也就不足为奇了。给他们时间。

was only accepted into C++ in Issaquah a few weeks ago (November 2016). It's not surprising that gcc or clang haven't implemented it yet. Give them time.

目前的解决方法是为 Overloader 创建线性层次结构。

The workaround for now is to create a linear hierarchy for Overloader instead.

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