because I want to overload all cv and reference qualifications of a member function, I wrote myself the following macro:
#define DEFINE_FUNCTION(sig , functionality) \
sig & { functionality; } \
sig && { functionality; } \
sig const & { functionality; } \
sig const && { functionality; } \
sig volatile & { functionality; } \
sig volatile && { functionality; } \
sig const volatile & { functionality; } \
sig const volatile && { functionality; }
Unfortunately, If I want to return *this
in a forwarding manner (that is preserving all reference and cv-qualification of the this
-pointer within the return type, it seems, I have to manually write all 8 of those overloads.
Now to my question:
Is it possible to get the cv and reference qualified type of *this
in a universal manner?
I have tried decltype(auto)
as return type and
return std::forward<decltype(*this)>(*this);
but apparently, the resulting expression always resolves to an lvalue reference, even in case of a &&
-qualified function.
Can you help me?
What you need is a friend.
struct blah {
decltype(auto) foo() & { return do_foo(*this); }
decltype(auto) foo() && { return do_foo(std::move(*this)); }
decltype(auto) foo() const& { return do_foo(*this); }
decltype(auto) foo() const&& { return do_foo(std::move(*this)); }
decltype(auto) foo() const volatile& { return do_foo(*this); }
decltype(auto) foo() const volatile&& { return do_foo(std::move(*this)); }
decltype(auto) foo() volatile& { return do_foo(*this); }
decltype(auto) foo() volatile&& { return do_foo(std::move(*this)); }
blah( const volatile blah&& b ) {}
blah( const volatile blah& b ) {}
blah( const blah& b ) = default;
blah( blah&& b ) = default;
blah() = default;
template<class Self>
friend blah do_foo(Self&& self) {
std::cout << "Is reference:" << std::is_reference<Self>::value << "\n";
std::cout << "Is const:" << std::is_const<std::remove_reference_t<Self>>::value << "\n";
std::cout << "Is volatile:" << std::is_volatile<std::remove_reference_t<Self>>::value << "\n";
return decltype(self)(self);
}
};
test code:
blah{}.foo();
blah tmp;
tmp.foo();
const blah tmp2;
tmp2.foo();
Output:
Is reference:0 Is const:0 Is volatile:0 Is reference:1 Is const:0 Is volatile:0 Is reference:1 Is const:1 Is volatile:0
Use the macro to implement the methods that forward to the friend, and write the friend manually. The methods that forward to the friend could even be template<class...Args>
if you are ok with the disadvantages of perfect forwarding.
Having less of your code that runs in a macro will be worth it.
#define RETURNS(...) \
noexcept(noexcept(__VA_ARGS__)) \
-> decltype( __VA_ARGS__ ) \
{ return __VA_ARGS__; }
#define FORWARD_METHOD_TO_FRIEND( NAME, SELF, REF_QUAL, HELPER_NAME ) \
template<class...Args>
auto NAME( Args&&... args ) REF_QUAL \
RETURNS( HELPER_NAME( SELF, std::forward<Args>(args)... ) )
#define FORWARD_CV_METHOD_TO_FRIEND( NAME, SELF, REF_QUAL, HELPER_NAME ) \
FORWARD_METHOD_TO_FRIEND( NAME, SELF, REF_QUAL, HELPER_NAME ) \
FORWARD_METHOD_TO_FRIEND( NAME, SELF, const REF_QUAL, HELPER_NAME ) \
FORWARD_METHOD_TO_FRIEND( NAME, SELF, const volatile REF_QUAL, HELPER_NAME ) \
FORWARD_METHOD_TO_FRIEND( NAME, SELF, volatile REF_QUAL, HELPER_NAME )
#define FORWARD_SELF_TO_FRIEND( NAME, HELPER_NAME ) \
FORWARD_CV_METHOD_TO_FRIEND( NAME, *this, &, HELPER_NAME ) \
FORWARD_CV_METHOD_TO_FRIEND( NAME, std::move(*this), &&, HELPER_NAME )
which shortens blah
above to:
struct blah {
FORWARD_SELF_TO_FRIEND( foo, do_foo )
// ...
template<class Self>
void do_foo( Self&& self )
{
// some_blah.foo() invokes do_foo passing it
// self with the proper qualifications
}
};
replacing (and improving) all of the foo()
methods.
some_blah.foo()
now invokes do_foo( some_blah )
with proper CV and L/R value qualifications and noexcept on some_blah
.
constexpr
may require some love to get right, because we don't have a conditional constexpr test in C++ right now. Maybe simply claiming constexpr
in the template foo
generated is the right thing, but I'm hazy on the subject.