Lambda with explicit object parameter and std::invoke_result_t

Question

How do I get the return type of a lambda that has a deducing this signature using std::invoke_result_t.

auto lambda = [](this auto& self, int x) -> int {
    return x;
};  
  
auto x = std::invoke_result_t<decltype(lambda), int>{}; //won't compile

Do I need to somehow specify the self argument inside the std::invoke_result_t ?

I've tried without "deducing this" and the above sample works.

Edit: compiler explorer link

I'm playing around with meta-programming a bit. I've tried passing std::invoke_result_t<decltype(lambda), decltype(lambda)&, int>{}; and similar. It didn't work for me. — Gallaros
– Gallaros, Commented Sep 5, 2023 at 7:29
Your std::invoke_result_t<decltype(lambda), decltype(lambda)&, int>{}; seems to work; EDIT but maybe because without __cpp_explicit_this_parameter — pptaszni
– pptaszni, Commented Sep 5, 2023 at 7:42
Your snippet works in MSVC latest (which has full support for C++23): godbolt.org/z/McGbGKzdY, but doesn't in any other compiler (i.e. GCC godbolt.org/z/3ox9roc14). I would say it is still an unsupported feature... — LoPiTaL
– LoPiTaL, Commented Sep 5, 2023 at 8:13
Sorry, my bad. Your snippet works with a slight modification to the this parameter. It works without the reference marker — LoPiTaL
– LoPiTaL, Commented Sep 5, 2023 at 8:15
But passing by reference (I think) is an error. this must be passed either by value (pointer copy) or as a forwarding reference. See devblogs.microsoft.com/cppblog/cpp23-deducing-this. Both work fine — LoPiTaL
– LoPiTaL, Commented Sep 5, 2023 at 8:18

康桓瑋 · Accepted Answer · 2023-09-05 08:42:49Z

7

In your lambda, this auto& self actually takes an lvalue reference to this, but std::invoke_result_t<decltype(lambda), int> invokes the rvalue lambda.

Since the rvalue cannot be bound to the lvalue reference, invoke_result has no valid member type, just as std::move(lambda)(0) is ill-formed.

You should invoke the lambda with an lvalue, e.g.

auto x = std::invoke_result_t<decltype(lambda)&, int>{};

Or make the lambda accept a forward reference to this

auto lambda = [](this auto&& self, int x) -> int {
    return x;
};

edited Sep 5, 2023 at 8:42

answered Sep 5, 2023 at 8:19

康桓瑋

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