Could not deduce template arguments (for a simple class function)

Asked Jul 04 '20 at 06:52

Active Jul 04 '20 at 06:52

Viewed 15 times

Suppose we have the following class function:

struct Entity {}

template <typename Component, typename... Components>
void each(std::function<void(Entity&, Component&, Components&...)> fn) {}

Calling it like below produces the "could not deduce template argument for..." error:

.each<int>([](Entity& e, int& i) {}

However, this works as expected:

std::function<void(Entity&, int&)> x = [](Entity& e, int& i) {}

.each<int>(x);

Straight forward, what's going on here and how to solve this?

asked Jul 04 '20 at 06:52

Yves Calaci

Why not just `template void each(Fn&& fn)`? Lambda is not `std::function` (but can be converted into one). – Evg Jul 04 '20 at 06:53
Because `.each` forwards the `std::function` to another function and wrong template params can be difficult and messy to deal with on the error log. – Yves Calaci Jul 04 '20 at 06:58
You can't deduce template parameters for `std::function` from something that is not `std::function`. You need one level of indirection. – Evg Jul 04 '20 at 07:00
So there's no way to achieve this other than your first comment? – Yves Calaci Jul 04 '20 at 07:05
I guess, no. First take `Fn&& fn`, then convert it into `std::function` and call your `each`. – Evg Jul 04 '20 at 07:06
How is this conversion done? One can input anything for `fn`: an simple integer, an object, a function pointer and so on. This is why I implicitly used `std::function` instead. – Yves Calaci Jul 04 '20 at 07:10
Do you have access to C++17? – Evg Jul 04 '20 at 07:20
Yes... abcdefgh – Yves Calaci Jul 04 '20 at 07:30
How about this https://godbolt.org/z/BwTa-z ? – Evg Jul 04 '20 at 07:33

0 Answers0