Can a noexcept function still call a function that throws in C++17?

Question

In P0012R1, "Make exception specifications be part of the type system",
I see that noexcept is now becoming a part of the function type.

I can't tell whether this will prevent noexcept(true) functions from still being able to call noexcept(false) functions.

Will the following code still be valid for C++17?

void will_throw() noexcept(false){
  throw 0;
}

void will_not_throw() noexcept(true){
  will_throw();
}

Answer 1

According to cppreference:

Note that a noexcept specification on a function is not a compile-time check; it is merely a method for a programmer to inform the compiler whether or not a function should throw exceptions.

So the syntax of your code is valid, but std::terminate will be called when executed.

Answer 2

noexcept(true) functions can call noexcept(false) functions. There will be a runtime error if an exception is thrown. The canonical example of why this is allowed is:

double hypotenuse(double opposite, double adjacent) noexcept(true)
{
    return std::sqrt(opposite*opposite + adjacent*adjacent);
}

std::sqrt will throw domain_error if its argument is negative, but clearly that will never happen here.

(In an ideal world, it would be forbidden by default with an exception_cast to allow it where required. The result could either be UB if an exception is thrown, or std::terminate).