It is possible to do such recursive template check, but it makes code difficult to read.
The principle is to forward recursive template check to a function found by dependent name look up, whose constraints will only be verified if the type does not already belong to a list of already checked types... If the type belong to the list of already checked type, the function is disabled by SFINAE, and an other function that does not recursively refers to the concept is selected by overload resolution:
See it in action: compiler-explorer-link
#include <type_traits>
namespace trying{
struct to_do{};
template <class...Checked, class T>
std::enable_if_t <(std::is_same_v <T,Checked> || ...), std::true_type>
too_complex(T &&, to_do);
template <class...Checked, class T>
std::false_type
too_complex(T &&,...);
}
template <class U, class T, class...Checked>
concept Integer_= requires(const T& a, const T& b, const U& to_be_readable)
{
requires decltype(too_complex <T, Checked...> (a + b, to_be_readable))::value ;
};
template <class T, class...Checked>
concept Integer = Integer_ <trying::to_do, T, Checked...>;
namespace trying{
template <class...Checked, class T>
requires (Integer <T, Checked...>)
std::enable_if_t <!(std::is_same_v <T,Checked> || ...), std::true_type>
too_complex(T &&, to_do);
}
struct x{
auto
operator + (x) const -> int;
};
struct y{
auto
operator + (y) const -> void*;
};
struct z2;
struct z1{
auto
operator + (z1) const -> z2;
};
struct z2{
auto
operator + (z2) const -> z1;
};
static_assert (Integer <int>);
static_assert (Integer <x>);
static_assert (!Integer <y>);
static_assert (Integer <z1>);
static_assert (Integer <z2>);
So yes it is possible... but I don't think it should be done.