c++ requires expression with inverse return type concept check

Question

I try to define a concept with a requires expression which checks some member function existence on the type.
The problem I run into is, that a single concept has to be plugged in for the return type check without any preceding bool operations like "not". This forces me to also write concepts for the inverse if needed, which is not really handy.
Is there something I am missing or another good way to do it?

Simple example:

template <typename T>
concept IsReference = std::is_reference_v<T>;

template <typename T>
concept HasSomeFunction = requires (T t) {
    {t.func()} -> IsReference; /* this is ok */
    {t.anotherFunc()} -> (not IsReference); /* this is not ok */
};

Answer 1

Here is a syntax that works on latest gcc and clang.

template <typename T>
concept HasSomeFunction = requires (T t) {
    {t.func()} -> IsReference; /* this is ok */
    requires !IsReference<decltype(t.anotherFunc())>;
};

Demo

It a bit harder to read. Maybe someone with some more experience with concepts can provide a better solution.

Answer 2

尝试这个

#include <memory>
#include <cassert>

struct B {};
struct D : B {};

template <typename T>
concept IsReference = std::is_reference_v<T>;
template <typename T>
concept NotIsReference = !std::is_reference_v<T>;

template <typename T>
concept HasSomeFunction = requires (T t) {
    {t.func()} -> IsReference; 
    {t.anotherFunc()} -> NotIsReference; 
};


int main() {
    std::shared_ptr<B> b = std::make_shared<D>();
    auto d = static_pointer_cast<D>(b);
    assert(d);
}