No-except without passing argument

Question

Can you say how can I assertion, is function noexcept(without passing arguments)? Thanks.

Answer 1

Assuming you are asking (thanks to @VittorioRomeo for the assumption):

How can I assert that a function is noexcept without calling it?

You can use the noexcept operator to do that, because its operands are unevaluated operands.
In other terms, you can do this:

void f(int) noexcept { }
void f(int, int) { }

int main() {
    static_assert(noexcept(f(0)), "!");
    static_assert(not noexcept(f(0, 0)), "!");
}

f won't be called in any case, that is the (let me say) nature of an unevaluated operand.

The most interesting part is that you can combine it with std::declval if you don't have a variable to be used as an argument and you don't know how to construct it.
As an example:

#include<utility>

struct S { S(int) {} };

void f(S) noexcept { }

int main() {
    static_assert(noexcept(f(std::declval<S>())), "!");
}

Note that I don't have to pass anything to construct S, even if it hasn't a default constructor.

This is usually enough to work around the requirement of passing no arguments.

Answer 2

Assuming you're asking:

How can I assert that a function is noexcept without calling it?

The answer is that you can't, because the noexcept specifier could be different between overloads. Example:

int a(int) noexcept;
int a(int, int);

// What would `noexcept(a)` mean?