C++ complicated lambda expressions

Question

Im doing a bit crazy project so dont get scared Im creating some kind of lambda calculus in CPP. Anyway how do I make it work?

auto INC_ = [=](void *arg, void *(*n)(void*, void*))-> void * {return n(arg,arg); };
auto do_ = [=](void *arg1, void *arg2)-> void * {return arg1; };
INC(a,do_);

INC_ gets 2 arg however second arg is a lambda exp with 2 other arguments.

do_ do what must be done with 2 arg

However if I try to call it all together like in 3 line. I get wrong arguments error.

So the real question is how to pass 2 arg lambda so it executes properly?

What is with all the void*s? C++ has a type system that is pretty useful. — NathanOliver
– NathanOliver, Commented May 9, 2017 at 18:07
I get wrong arguments error. And, exact error text is.. What, exactly? — Algirdas Preidžius
– Algirdas Preidžius, Commented May 9, 2017 at 18:08
Im required to get as close as possible to typeless code. I wrote at the beginning that its a bit crazy ;) If you can show example on int i can try to copy it into fancy void* — Gryfit
– Gryfit, Commented May 9, 2017 at 18:10
Ah. If you want *typeless" code you should use generic code (templates). See this reworked example: coliru.stacked-crooked.com/a/7d3d1c246a9a0564 — NathanOliver
– NathanOliver, Commented May 9, 2017 at 18:12
"Function cant be called with the given arg list" ;) And more after comp: error C2064: term does not evaluate to a function taking 2 arguments note: class does not define an 'operator()' or a user defined conversion operator to a pointer-to-function or reference-to-function that takes appropriate number of arguments — Gryfit
– Gryfit, Commented May 9, 2017 at 18:12

Yakk - Adam Nevraumont · Accepted Answer · 2017-05-09 18:16:05Z

1

A lambda is not a function pointer.

auto INC_ = [=](void *arg, void *(*n)(void*, void*))-> void * {return n(arg,arg); };
auto do_ = [=](void *arg1, void *arg2)-> void * {return arg1; };
INC(a,do_);

The second argument to INC_ is a function pointer.

A stateless lambda can be converted to a function pointer with compatible signature.

Once there is state, this is not possible. [=] makes it stateful, even though it doesn't capture anything.

In addition, don't use function pointers in typeless code.

Try this:

int a = 0;
auto INC_ = [](auto arg, auto n) {return n(arg,arg); };
auto do_ = [](auto arg1, auto arg2) {return arg1; };
std::cout << INC(a,do_) << "\n";

live example.

answered May 9, 2017 at 18:16

Yakk - Adam Nevraumont

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