How to deduce if a type is lambda?

Question

how to deduce a type is lambda ? like std::is_function ? I've also tried std::is_invocable, but it need exactly types of parameters.

You can check whether it is invokable with specific parameter types. There are arcane techniques for checking whether it has *any* `operator()`, but this is usually not a valid thing to check. In any case, you can't check whether it's specifically a lambda type. — Brian Bi, Oct 02 '21 at 00:37
@BrianBi, I think that is sad but true :(. I tried many ways... — VeNToR, Oct 02 '21 at 00:41
It's just an indication that you need to rethink the broader problem you are trying to solve, and find another way that doesn't depend on knowing whether a type is a lambda. — Brian Bi, Oct 02 '21 at 00:43
Lambdas *are not special* in C++. There is never any reason to write a function that can *only* take a lambda. Functions should accept any particular callable object appropriate to its use. — Nicol Bolas, Oct 02 '21 at 00:43
@BrianBi: Yeah, by my understanding a lambda is syntactic sugar for a standard functor class which a user could write by hand (the captures are constructor arguments, the arguments are the arguments to `operator()`); at best, with RTTI, you might be able to do terrible heuristics to figure out if the underlying name is something the cat spat up (probably a compiler generated name for the lambda) vs. a vaguely human name (though with the name-mangling compilers apply, it's going to look a lot like cat sick too). But even if you can, and even if heuristics work, there's never a reason to do it. — ShadowRanger, Oct 02 '21 at 00:44
@VeNToR: I strongly suspect this is [an XY problem](https://meta.stackexchange.com/q/66377/322040). *Why* do you think you need to do this? What problem are you solving? — ShadowRanger, Oct 02 '21 at 00:46
@ShadowRanger: my code is very long to share it. I wish I could share ? — VeNToR, Oct 02 '21 at 00:49
@VeNToR: I'm not asking for all your code. Just an explanation for why you think it's important to *specifically* identify lambdas and handle them differently from other functors, which they are drop-in equivalents for. The explanation could lead to an answer that solves your actual problem; the whole point of XY problems is that you've settled on a bad solution to your real problem, and you're trying to make the bad solution work, neglecting to describe the real problem that has a good solution. — ShadowRanger, Oct 02 '21 at 00:52
With https://stackoverflow.com/a/56766138/2193968 you can do https://godbolt.org/z/G9dWrMaW4 — Jerry Jeremiah, Oct 03 '21 at 22:52
@VeNToR Well, since string_view doesn't have contains() until C++23 what about this: https://godbolt.org/z/d39Ed69nW — Jerry Jeremiah, Oct 05 '21 at 02:08

score 2 · Answer 1 · answered Oct 02 '21 at 00:47

You cannot check for every invocable types without having the exact parameters. Take this for example:

auto my_lambda = [](std::integral auto number) {};

There are few properties that make it hard to check without knowing the parameter types:

You cannot take the address of operator() since it's a template. You could take the address of operator()<int> but you'd have to know the type in advance
You cannot try the lambda with a "convert to anything" type, since it's not an integral
There are no other observable properties of the type that can be observed.

That makes it really hard to have a is_lambda trait. Also, you could have something like that:

struct A {
    void operator()(int) const {}
};

This would be indistinguishable from a lambda.

There are however, very arcane ways to check if a type is a lambda or not.

I would strongly recommend not to do that.

You could theorically check the name of the type by using __PRETTY_FUNCTION or __FUNCSIG__, and check if the name look like a compiler generated name for a lambda, but that would exclude any other user defined callable types, which are, and should be accepted like lambda.

I leave the implementation as an exercise to the reader.

How to deduce if a type is lambda?

1 Answers1