我想要某种方法来获取 lambda 函数的第一个参数类型,这可能吗?
例如
代替:
template<typename T>
struct base
{
virtual bool operator()(T) = 0;
}
template<typename F, typename T>
struct filter : public base<T>
{
virtual bool operator()(T) override {return /*...*/ }
};
template<typename T, typename F>
filter<T> make_filter(F func)
{
return filter<F, T>(std::move(func));
}
auto f = make_filter<int>([](int n){return n % 2 == 0;});
我想:
template<typename F>
struct filter : public base<typename param1<F>::type>
{
bool operator()(typename param1<F>::type){return /*...*/ }
};
template<typename F>
filter<F> make_filter(F func)
{
return filter<F>(std::move(func));
}
auto f = make_filter([](int n){return n % 2 == 0;});
根据 Xeo 的回答,这就是我在 VS2010 中所做的工作:
template<typename FPtr>
struct arg1_traits_impl;
template<typename R, typename C, typename A1>
struct arg1_traits_impl<R (C::*)(A1)>{typedef A1 arg1_type;};
template<typename R, typename C, typename A1>
struct arg1_traits_impl<R (C::*)(A1) const>{typedef A1 arg1_type;};
template<typename T>
typename arg1_traits_impl<T>::arg1_type arg1_type_helper(T);
template<typename F>
struct filter : public base<typename std::decay<decltype(detail::arg1_type_helper(&F::operator()))>::type>
{
bool operator()(typename std::decay<decltype(detail::arg1_type_helper(&F::operator()))>::type){return /*...*/ }
};
template<typename T, typename F>
filter<F> make_filter(F func)
{
return filter<F>(std::move(func));
}
我试过简化代码,但任何尝试似乎都会破坏它。