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I'm trying to provide a hash<> specialization for a family of types I've been working on. So far so good, the specialization itself is easy to provide, I've already done similar things for numeric_limits<>. But I'm facing the problem of how to provide the specialization in a way that is portable to C++11 and -pre-11 (C++03 or whatever).

The problem I'm bumping on, of course, is that hash<> can be defined in one of several namespaced, and I'm forced to provide the specialization in the same namespace.

// in c++11
namespace std { struct hash<>...; }
// in c++03 this is one possibility
namespace std { namespace tr1 { struct hash<>...; } }

// my library's specialization
namespace ????_open {
struct hash<mytype>...;
????_close

Of course, one option would be using #defines to reach, open and close the adequate namespaces, or providing N files with N different specializations and conditionally #include the correct one, but that's cumbersome:

#if defined(some_c++11_condition)
  #include "c++11-specialization.h"
#elif defined(some_c++03_condition)
  #include "c++03-specialization.h"
#elif (some_other_condition)
  #oh_dear_who_knows_what_this_include_will_be_like
#else_ad_nauseam
#endif

Of course, I'll be sticking to that strategy if I'm forced to but I wanted to explore some other options before. In particular, I though I could use namespace alias to specialize in the correct place:

#if defined(some_c++11_condition)
  namespace std_specialize = std;
#elif defined(some_c++03_condition)
  namespace std_specialize = std::tr1;
#...
#endif

...

namespace std_specialize {
struct hash<mytype>...;
}

Unfortunately this doesn't work in any of the 3 compilers I have tried (MSVC 2008, GCC 4.7, Clang 3.0) with various errors regarding "declaration of namespace conflicts with..." in the line that reopens the namespace, which shouldn't happen because namespaces can be reopened multiple times, and if an alias is an alias and not something else then this should also apply for them.

So, is a namespace alias really an alias, or a misnomer meaning something else? Or is there other reason why I can not specialize this way? If so, any other method (better than barrages of #defines)?

Luis Machuca
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2 Answers2

1

Yes, a namespace alias is really an alias. The error is occurring because you're declaring std_specialize as an alias for either std or std::tr1 and then trying to declare a namespace with the same name. If it were legal, after that declaration, what would std_specialize refer to, std(or std::tr1) or the new namespace containing your hash specialization?

What you probably want to do is

namespace std {
#if __cplusplus < 201103L
namespace tr1 {
#endif

template<>
struct hash<my_type> { ... };

#if __cplusplus < 201103L
}
#endif
}
Praetorian
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  • Is it really trying to _declare_ a new namespace? The namespace `std_specialize` already exists, as an alias to another one, so the declaration should be reopening the existing namespace. Is that case not considered in the Standard, a parsing issue or something? – Luis Machuca Sep 26 '13 at 20:17
  • @Luis I realize that's what you want it to mean, but that's not what it's actually doing :). You are trying to declare a new namespace in that case. I don't know what the technical reason for disallowing your interpretation is. – Praetorian Sep 26 '13 at 20:23
0

Proper macros:

// Defined before
// #define NOEXCEPT noexcept

#include <iostream>

#if TR1_CONDITION
// tr1
    #define HASH_NAMESPACE std::tr1
    #define HASH_NAMESPACE_BEGIN namespace std { namespace tr1 {
    #define HASH_NAMESPACE_END }}
#else
// std
    #define HASH_NAMESPACE std
    #define HASH_NAMESPACE_BEGIN namespace std {
    #define HASH_NAMESPACE_END }
#endif

namespace X {
    struct Class {};
}

HASH_NAMESPACE_BEGIN
    template<>
    struct hash<X::Class>  {
        size_t operator()(const X::Class&) const NOEXCEPT { return 123; }
    };
HASH_NAMESPACE_END

int main() {
    HASH_NAMESPACE::hash<X::Class> h;
    std::cout << h(X::Class()) << std::endl;
    return 0;
}

(But all of it is not nice)