Is it legal to use a const value captured in a lambda as a template argument?

Question

Consider the following code, provided by a colleague:

#include <array>
#include <string>

int main() {
    const int size = 4;
    return [size]() {
      std::array<std::string, size> a; // *
      return a.size();
    }();
}

It's accepted by Clang 5.0.0 but rejected by GCC 7.2 with the error message for the starred line being:

error: '__closure' is not a constant expression

Which compiler is right?

Answer 1

The rule is actually intuitive: any occurrence of a variable that doesn't necessitate a capture refers to the original variable. [expr.prim.lambda]/11:

Every id-expression within the compound-statement of a lambda-expression that is an odr-use of an entity captured by copy is transformed into an access to the corresponding unnamed data member of the closure type. [ Note: An id-expression that is not an odr-use refers to the original entity, never to a member of the closure type. […] — end note ]

Clearly, the declared size variable can be used in constant expressions, hence Clang is right.