Lambda functions as base classes

Question

Playing around with Lambdas I found an interesting behaviour that I do not fully understand.

Supose I have a struct Overload that derives from 2 templat

Answer 1

In addition to operator(), a the class defined by a lambda can (under the right circumstances) provide a conversion to a pointer to function. The circumstance (or at least the primary one) is that the lambda can't capture anything.

If you add a capture:

auto f1 = get(
              []() { std::cout << "lambda1::operator()()\n"; },
              [i](int) { std::cout << "lambda2::operator()(int)\n"; }
              );
f1();
f1(2);

...the conversion to pointer to function is no longer provided, so trying to compile the code above gives the error you probably expected all along:

trash9.cpp: In function 'int main(int, char**)':
trash9.cpp:49:9: error: no match for call to '(Overload, main(int, char**):: >) (int)'
trash9.cpp:14:8: note: candidate is:
trash9.cpp:45:23: note: main(int, char**)::
trash9.cpp:45:23: note:   candidate expects 0 arguments, 1 provided