Will an exception thrown from a noexcept function parameter's constructor immediately result in a call to std::terminate()?

Question

Given the following class declaration:

class phone_number
{
public:
    explicit phone_number( std::string number ) noexcept( std::is_nothrow_move_constructible< std::string >::value );
}

phone_number::phone_number( std::string number ) noexcept( std::is_nothrow_move_constructible< std::string >::value )
    : m_originalNumber{ std::move( number ) }
{

}

Will the following line of code end up calling std::terminate() immediately due to the noexcept specification if an exception is thrown from the string constructor?

const phone_number phone("(123) 456-7890");

Answer 1

Since all the parameters are evaluated before the function is invoked, an exception, emitted by a parameter's constructor, would not violate noexcept contract of the function itself.

To confirm this, here's what I've tried, approximating your example:

class A
{
public:
    A(const char *)
    {
        throw std::exception();
    }
};

void f(A a) noexcept
{

}

int main()
{   
    try
    {
        f("hello");
    }
    catch(std::exception&)
    {
        cerr<< "Fizz..." << endl;
    }
    return 0;
}

The output, unsurprisingly, was Fizz... and the program exited normally.