For this code:

struct A {};
struct B {};

auto f() -> std::variant<A, B>
{
    if (true) return A();
    return B();
}

Is it possible for a compiler to automatically deduce return type as std::variant<A,B>?
Just so return type declaration can be omitted:

auto f() // std::variant<A, B> deduced by compiler
{
    if (true) return A();
    return B();
}