What is the type of a "variable" containing a lambda function that captures by reference?

Question

I was doing a quick experiment with lambda functions and I'm having trouble figuring out how to declare the testFunc variable below.

Generally speaking, a function pointer can be defined as follows:

int (*someFunc)(int, int) = otherFunction;

When doing this for a lambda function without any closures, it works fine:

int (*someFunc)(int) = [](int a) -> int { return 0; };

However, I get a compiler error when trying to capture by reference:

void (*testFunc)() = [&]() -> void { /* code here */ };

It works if I declare testfunc as auto instead, but I'm curious what's wrong with the above code?

you can use `std::function` but should be aware of overhead, it is pretty small but exists. — Slava, Jan 09 '18 at 22:36

score 2 · Accepted Answer · edited Jan 09 '18 at 22:00

2

It's pretty simple:

All lambdas have unique anonymous types. (that's why auto works)

Lambdas without captures are convertible to function pointers, but capturing lambdas are not.

edited Jan 09 '18 at 22:00

Remy Lebeau

answered Jan 09 '18 at 21:50

HolyBlackCat

Excellent answer! Could you give an example definition of a variable containing a lambda with a capture closure? I'm guessing it requires a template class? – Gogeta70 Jan 09 '18 at 22:03
@Gogeta70 I'm not sure what you mean by 'capture closure'. – HolyBlackCat Jan 09 '18 at 22:10
I mean how would I define `testFunc` above in my question? – Gogeta70 Jan 09 '18 at 22:11
2

@Gogeta70 It is not possible without `auto`. Lambdas capturing state can not be converted to regular function pointer. – user7860670 Jan 09 '18 at 22:13
Lambda types are called anonymous for a reason. :) – HolyBlackCat Jan 09 '18 at 22:14
You can assign them to `std::function` though – Slava Jan 09 '18 at 22:33

1 Answers1