Trouble understanding C++14 Relaxed constexpr restrictions

Question

I've came across new C++14 signature for std::max function:

template< class T > 
const T& max( const T& a, const T& b ); // (C++11)

template< class T > 
constexpr const T& max( const T& a, const T& b );// (C++14)

I've read about Relaxed constexpr restrictions proposal for C++14 but I'm still don't understand why this function return value can be constexpr

Example:

std::vector<int> a, b;
//This does not compile but as my understadnding of `constexpr` this should
int array[std::max(a.size(), b.size()]; (1)
//This is trivial use that does compile
int array[std::max(1,2)]; (2)

When calling std::max in (1) constexpr is ignored ?

Answer 1

The base issue isn't directly related to relaxed constexpr rules, a constexpr function is only a constant expression if the arguments are constant expressions. In your second case:

int array[std::max(1,2)];

this works because integer literals are indeed constant expressions.

C++11 was more specific in this case, the draft C++11 standard in section 5.19 [expr.const] lays out the cases where a sub-expression is not considered a constant expression and contains the following:

an invocation of a constexpr function with arguments that, when substituted by function invocation substitution (7.1.5), do not produce a constant expression;

in C++14 this paragraph was removed and we have the following exception:

an invocation of a function other than a constexpr constructor for a literal class, a constexpr function, or an implicit invocation of a trivial destructor (12.4) [ Note: Overload resolution (13.3) is applied as usual —end note ];

and we have to use the rest of the rules with respect to the arguments(sub-expressions).

In the first case:

int array[std::max(a.size(), b.size()];

std::vector::size is not marked constexpr and so it falls under the above quoted exception.

Also note that in section 7.1.5 [dcl.constexpr] we have the following:

A call to a constexpr function produces the same result as a call to an equivalent non-constexpr function in all respects except that a call to a constexpr function can appear in a constant expression.

Passing arguments to a constexpr function that are not constant expressions just means the expression is not available for uses in contexts that require a constant expression.

As dyp points out that std::max was not made constexpr in C++11 due to various issues probably including committee being conservative and limited time, we can see some of the issues involved in N3039. it being forgotten see N3856 which says:

This short paper proposes to make the standard functions min and max constexpr. They were top on the list of motivating cases for al- lowing reference parameters for constexpr functions in C++11. They were forgotten after the core language change was accepted

For reference we know that 1 and 2 are constant expression from section 5.19 which says:

A literal constant expression is a prvalue core constant expression of literal type, but not pointer type. An integral constant expression is a literal constant expression of integral or unscoped enumeration type. [...]