泛型 lambda 的数量
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.
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