c++ - warning: specialization of template in different namespace

Question

With the following piece of code I get the warning:

warning: specialization of ‘template<class _Iterator> struct std::iterator_traits’ in different namespace [-fpermissive]

template<> class std::iterator_traits<Token_ptr>{
public:
    typedef Word difference_type;
    typedef Word value_type;
    typedef Token_ptr pointer;
    typedef Word& reference ;
    typedef std::bidirectional_iterator_tag iterator_category ;
};

While everything works correctly, does any body know what exactly means and why is issued the warning. ( g++ issues the warning while clang++ doesn't ).

Answer 1

Assuming that you are compiling this in C++11 mode (since clang gave no warning) and that this specialization is in the global namespace, then there's nothing wrong with your code. It's a g++ bug. §14.7.3 [temp.expl.spec]/p2:

An explicit specialization shall be declared in a namespace enclosing the specialized template. An explicit specialization whose declarator-id is not qualified shall be declared in the nearest enclosing namespace of the template, or, if the namespace is inline (7.3.1), any namespace from its enclosing namespace set. Such a declaration may also be a definition. If the declaration is not a definition, the specialization may be defined later (7.3.1.2).

The global namespace is a "namespace enclosing the specialized template", and your declarator-id is qualified with std::, so the second sentence doesn't apply. As a workaround, you can do what cdhowie's answer suggests - i.e., opening a namespace std block and putting the specialization there.

See CWG issue 374 and GCC bug 56480.