CODEWITHSUNDEEP
c++c++11
Candy Chiu
Candy Chiu

Reputation: 6679

How to interpret declval<_Dest>() = declval<_Src>() in is_assignable

I am trying to figure out how to interpret declval<_Dest>() = declval<_Src>() in the implementation of is_assignable.

declval turns a type into a reference. Given that, I translate the expression into one of the following four possibilities:

  1. _Dest&& = _Src&&
  2. _Dest&& = _Src&
  3. _Dest& = _Src&&
  4. _Dest& = _Src&

I then created two helper functions.

template <typename T> T rvalue();
template <typename T> T& lvalue();

My understanding is the four expressions can be realized by using the template functions.

  1. _Dest&& = _Src&& -----> rvalue<_Dest>() = rvalue<_Src>()

Same goes for the other three.

Then I simulated decltype(declval<_Dest>() = declval<_Src>(), ..) by compiling the templated function version of each of the possibilities for three pairs of concrete types.

My questions are

Thanks.

Upvotes: 3

Views: 232

Answers (1)

Casey
Casey

Reputation: 42554

std::declval is actually specified to be (C++11 §20.2.4 [declval] p1):

template <class T>
typename add_rvalue_reference<T>::type declval() noexcept;

The result of the reference collapsing rules (§8.3.2 [dcl.ref] p6) is that declval returns an lvalue reference when T is an lvalue reference type, and an rvalue reference otherwise. So yes, your interpretation is correct.

If your compiler thinks that double&& is assignable from any type, then it has a bug. §5.17 [expr.ass] p1 states:

The assignment operator (=) and the compound assignment operators all group right-to-left. All require a modifiable lvalue as their left operand and return an lvalue referring to the left operand.

[emphasis mine].

Many programmers choose to emulate this behavior - assingment only to lvalues - with their own types by declaring the assignment operators with an lvalue reference qualifier:

class foo {
  foo& operator = (const foo&) & = default;
  foo& operator = (foo&&) & = default;
};

Upvotes: 2

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