I am trying to understand how std::declval<T>()
works. I know how to use it, and know what it does, mainly allows you to use decltype
without constructing the object, like
decltype(std::declval<Foo>().some_func()) my_type; // no construction of Foo
I know from cppreference.com that std::declval<Foo>
"adds" a rvalue reference to Foo
, which due to reference collapsing rules ends up being either a rvalue reference or a lvalue reference. My question is why the constructor of Foo
is not called? How can one implement a "toy" version of std::declval<T>
without constructing the template parameter?
PS: I know it is not the same as the old trick
(*(T*)(nullptr))
Basically, in a
sizeof
ordecltype
expression you can call functions that aren't implemented anywhere (they need to be declared, not implemented).E.g.
The linker should not complain about that.