alter alter - 1 year ago 65
C++ Question

how is this sizeof expression evaluated? and why it is called that way?

I came across with this code in an

implementation :

template <class T, class U>
struct is_assignable
template <class X, class Y>
constexpr static bool has_assign(...) { return false; }

template <class X, class Y, size_t S = sizeof((std::declval<X>() = std::declval<Y>(), true)) >
// the comma operator is necessary for the cases where operator= returns void
constexpr static bool has_assign(bool) { return true; }

constexpr static bool value = has_assign<T, U>(true);

The part that I cant understand how it works or how it is evaluated is
size_t S = sizeof((std::declval<X>() = std::declval<Y>(), true))
I know that if the assign operation fails it will fall back to the first definition of has_assign that returns false, but i don't know why it has the
, true)

I did some test with structs that returns void on the assign operator and removing the
, true
part in
gives me the same results.

Answer Source

In order to apply sizeof(), you need a complete type. But returning a complete type isn't a requirement of assignability, hence:

sizeof((std::declval<X>() = std::declval<Y>(), true))
       ~~~~~~~~~~~~~~~~~~ expr ~~~~~~~~~~~~~~~~~~~~~

if the assignment is valid for those two types, then we have sizeof(expr) where the type of expr is bool (because true). So if the assignment is valid, we get some real size. Otherwise, substitution failure.

But this is an unnecessarily cryptic way of writing this code. Moreover, it's not even correct because I could write a type like:

struct Evil {
    template <class T> Evil operator=(T&& ); // assignable from anything
    void operator,(bool);                    // mwahahaha

and now your sizeof() still doesn't work.

Instead, prefer simply:

class = decltype(std::declval<X>() = std::declval<Y>())

This accomplishes the same result - either substitution failure or not - without needed to care at all about what the type of the result is or to handle special cases.

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