I have a generic lambda function that needs to accept a pointer-to-member-function as a parameter. I can of course simply use auto on its own and the compiler will deduce the correct type. However, where possible, I prefer to decorate my auto parameters with *, & and const where appropriate, thus better communicating the nature and intent of the deduced type. If I simply make the auto parameter an auto*, I get a compiler error, which I'm not really surprised by as auto* signifies a regular pointer, not a pointer-to-member. Is there some syntax for constraining an auto parameter to accept a pointer-to-member, or should I just use auto and forget about it?
int main()
{
struct S { void m() {} };
//auto l = [](auto* pmf) // ERROR
//auto l = [](const auto& pmf) // Works, but uh, bit misleading I think
auto l = [](auto pmf)
{
S s;
(s.*pmf)();
};
l(&S::m);
}
You can declare it as:
auto l = [](auto S::*pmf)
It does tie the pointer to a S type, but it makes sense because it is that way you will use it.
In C++20 you can constrain it with a concept:
#include <type_traits>
template <typename T>
concept MemberPointer = std::is_member_pointer_v<T>;
void test() {
auto foo = [](MemberPointer auto memPtr) {};
}