GCC 4.7.2 compiles this:
constexpr int i = 5;
[]{ std::integral_constant< int, i >(); }; // nonstandard: i not captured
but not this:
constexpr int i = 5;
[&i]{ std::integral_constant< int, i >(); }; // GCC says i not constexpr
The latter example appears correct to me, according to C++11 §5.1.2/15:
An entity is captured by reference if it is implicitly or explicitly captured but not captured by copy. It is unspecified whether additional unnamed non-static data members are declared in the closure type for entities captured by reference.
It seems the captured object i inside the lambda refers to the variable in the enclosing scope, which is constexpr, not merely a const reference.
The standard explicitly says that the use of a by-value capture is transformed into a use of the corresponding member of the lambda object. And I think that 5.1.2 hints that my interpretation is correct.
Is there anything that explicitly says that whether a capture by reference refers to the object in the enclosing scope or a reference?
The second template-argument to
std::integral_constant< int, i >is for a template-parameter of non-type form, specifically of integral or enumeration type (14.3.2p1 bullet 1) and so must be a converted constant expression of typeint.In a lambda-expression, implicit capture occurs when an entity is odr-used in the compound statement (5.1.2p11); use of a converted constant expression in an explicit template instantiation is not odr-use (3.2p3), so the first example is valid.
In the second example, I think gcc is incorrect to reject it; 5.1.2p17 says in a note that:
Although the paragraph as a whole is discussing capture by copy, there’s no reason not to apply this rule to capture by reference as well. It’s unsurprising that the standard is unclear on this; there’s really no reason to capture an entity that can be used in a converted constant expression by reference.