How to avoid splitting namespace scope?

[Jayden Shui]

Hello All,

I give every class a tag by specialize the following template
function:

template<class T>
std::string Tag();

However, when working on a class in a namespace such as

namespace S
{
class A { /*codes for A*/ };

template<>
std::string Tag<A>()
{
return "A";
}

// other codes
}

The compiler give an error. I have to specialize it in global scope
like

namespace S
{
class A { /*codes for A*/ };
}

template<>
std::string Tag<S::A>()
{
return "A";
}

namespace S
{
// other codes
}

I don't want split the namespace scope by the tag specialization.
Anyway I can make it work without the splitting?

Thank you very much for your kind help!

Best regards,

Jayden

 

[laird]

If all you want is a string describing the class, have you considered using the typeid operator?

S::A a;
cout << typeid(a).name() << endl;

or create a single global Tag function template:

template<class T> string Tag(T& t) { return typeid(a).name(); }

Laird Breyer.

 

[Jayden Shui]

If all you want is a string describing the class, have you considered using the typeid operator?

S::A a;
cout << typeid(a).name() << endl;

or create a single global Tag function template:

template<class T> string Tag(T& t) { return typeid(a).name(); }

Laird Breyer.

Thanks. But I'd like the tag more readable. Any other way?

Best regards,

Jayden

 

[laird]

The name() string's format depends on your compiler.

With Visual C++ the string should be something like "class S::A".

With GCC, the name is mangled like "N1S1AE" and the string can be demangled using c++filt

$ c++filt -t N1S1AE
S::A

See also
http://gcc.gnu.org/onlinedocs/libstdc++/manual/bk01pt12ch39.html

Laird.

 

[Bo Persson]

Hello All,

I give every class a tag by specialize the following template
function:

template<class T>
std::string Tag();

However, when working on a class in a namespace such as

namespace S
{
class A { /*codes for A*/ };

template<>
std::string Tag<A>()
{
return "A";
}

// other codes
}

The compiler give an error. I have to specialize it in global scope
like

namespace S
{
class A { /*codes for A*/ };
}

template<>
std::string Tag<S::A>()
{
return "A";
}

namespace S
{
// other codes
}

I don't want split the namespace scope by the tag specialization.
Anyway I can make it work without the splitting?

No really. You cannot specialize a template from one namespace inside
another namespace.


Bo Persson