The specification of std::addressof
was changed for C++17: it is now allowed to be a constant expression. However, cppreference says that:
The expression
std::addressof(E)
is a constant subexpression, ifE
is an lvalue constant subexpression.
std::addressof(E)
will be a constant expression?std::addressof(E)
will NOT be a constant expression?This is explained here.
Introduce the following new definition to the existing list in 17.3 [definitions]: [Drafting note: If LWG 2234 is accepted before this issue, the accepted wording for the new definition should be used instead — end drafting note]
**constant subexpression** [defns.const.subexpr] an expression whose evaluation as a subexpression of a *conditional-expression* *CE* (5.16 [expr.cond]) would not prevent *CE* from being a core constant expression (5.20 [expr.const]).
So "constant subexpression" roughly means "you can use it in a constant expression".
What is an example where std::addressof(E) will be a constant expression?
I believe it's intended to give a constant expression whenever &E
does (assuming that &
invokes the built-in address-of operator).
constexpr int x = 42; // static storage duration
constexpr int* p1 = &x; // x is an lvalue constant subexpression
constexpr int* p2 = std::addressof(x); // x is an lvalue constant subexpression
What is an example where std::addressof(E) will NOT be a constant expression?
std::map<int, int> m;
void f() {
int& r = m[42];
constexpr int* z1 = &r; // error: r is not a constant subexpression
constexpr int* z2 = std::addressof(r); // likewise
constexpr int x = 43; // automatic storage duration
constexpr const int y1 = *&x; // ok; x is a constant subexpression
constexpr const int y2 = *std::addressof(x); // likewise
constexpr const int* p1 = &x; // error: p1 points to an automatic object
constexpr const int* p2 = std::addressof(x); // likewise
}