c++ - Error using a constexpr as a template parameter within the same class

Question

If I try to compile the following C++0x code, I get an error:

template<int n> struct foo { };

struct bar {
    static constexpr int number() { return 256; }

    void function(foo<number()> &);
};

With gcc 4.6.1, the error message is:

test.cc:6:27: error: ‘static constexpr int bar::number()’ used before its definition
test.cc:6:28: note: in template argument for type ‘int’

With clang 2.8, the error message is:

test.cc:6:20: error: non-type template argument of type 'int' is not an integral
      constant expression
        void function(foo<number()> &);
                          ^~~~~~~~
1 error generated.

If I move the constexpr function to a base class, it works on gcc, and gives the same error message on clang:

template<int n> struct foo { };

struct base {
    static constexpr int number() { return 256; }
};

struct bar : base {
    void function(foo<number()> &);
};

Is the code wrong, or is it a limitation or bug on gcc 4.6's implementation of C++0x? If the code is wrong, why is it wrong, and which clauses of the C++11 standard say it is incorrect?

Answer 1

In C++, inline definitions of member functions for a class are only parsed after every declaration in the class is parsed. Therefore, in your first example, the compiler can't see the definition of number() at the point where function() is declared.

(No released version of clang has support for evaluating constexpr functions, so none of your testcases will work there.)