Why is swapping multidimensional arrays not noexce

问题:

I have the following snippet:

#include <algorithm>
#include <iostream>

int main(int argc, char** argv) {
    int x[2][3];
    int y[2][3];

    using std::swap;
    std::cout << noexcept(swap(x, y)) << "\n";

    return 0;
}

Using GCC 4.9.0, this prints 0. I don't understand why.

According to the standard there's two overloads for std::swap:

namespace std {
    template<class T> void swap(T& a, T& b) noexcept(
        is_nothrow_move_constructible<T>::value &&
        is_nothrow_move_assignable<T>::value
    );

    template<class T, size_t N>
    void swap(T (&a)[N], T (&b)[N]) noexcept(noexcept(swap(*a, *b)));
}

In my understanding the noexcept specifier for the array should work recursively for multidimensional arrays.

Why is swapping multidimensional arrays not noexcept?

---

While trying to find a minimal example that still behaves weirdly I came up with the following:

#include <iostream>

template<class T> struct Specialized      : std::false_type {};
template<>        struct Specialized<int> : std::true_type  {};

template<class T>                void f(T& a) noexcept(Specialized<T>::value);
template<class T, std::size_t N> void f(T (&a)[N]) noexcept(noexcept(f(*a)));

int main(int argc, char** argv) {
    int x, y[1], z[1][1];

    std::cout << noexcept(f(x)) << " "
              << noexcept(f(y)) << " "
              << noexcept(f(z)) << "\n";
}

Using GCC 4.9.0 this prints 1 1 0, but again I don't understand why.

回答1:

This overload:

template<class T, size_t N>
void swap(T (&a)[N], T (&b)[N]) noexcept(noexcept(swap(*a, *b)));

is not in scope until ;, so swap(*a, *b) doesn't consider this overload. This is because of:

3.3.2/1 The point of declaration for a name is immediately after its complete declarator (Clause 8) and before its initializer (if any) …

and the exception specification is part of the declarator.