问题描述

可以通过访问它的 operator() 来推断非泛型 lambda 的 arity.

It is possible to deduce arity of a non-generic lambda by accessing its operator().

template <typename F>
struct fInfo : fInfo<decltype(&F::operator())> { };

template <typename F, typename Ret, typename... Args>
struct fInfo<Ret(F::*)(Args...)const> { static const int arity = sizeof...(Args); };

这对于 [](int x){ return x;} 因为 operator() 不是模板化的.

This is nice and dandy for something like [](int x){ return x; } as the operator() is not templated.

但是，泛型 lambda 对 operator() 进行模板化，并且只能访问模板的具体实例 - 这有点问题，因为我无法手动为 operator() 提供模板参数code>operator() 因为我不知道它的 arity 是什么.

However, generic lambdas do template the operator() and it is only possible to access a concrete instantiation of the template - which is slightly problematic because I can't manually provide template arguments for the operator() as I don't know what its arity is.

所以，当然，像

auto lambda = [](auto x){ return x; };
auto arity = fInfo<decltype(lambda)>::arity;

不起作用.

我不知道要转换为什么，也不知道要提供哪些模板参数(或提供多少)(operator()).
任何想法如何做到这一点?

I don't know what to cast to nor do I know what template arguments to provide (or how many) (operator()<??>).
Any ideas how to do this?

推荐答案

这是不可能的，因为函数调用运算符可以是可变参数模板.一般来说，对于函数对象永远不可能这样做，特殊情况下的 lambdas 因为它们碰巧不是同样强大，所以总是一个坏主意.现在是时候让这个坏主意回家了.

It's impossible, as the function call operator can be a variadic template. It's been impossible to do this forever for function objects in general, and special-casing lambdas because they happened to not be equally powerful was always going to be a bad idea. Now it's just time for that bad idea to come home to roost.

这篇关于泛型 lambda 的数量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！