How can I specialize typedef and its implicit type in different ways?

Question

I have something like this:

typedef int AnotherType; template <typename T> Func( T Value ); // And I want to specialize these two cases separately: template <> bool Func<int>( int Value ) {...} template <> bool Func<AnotherType>( AnotherType Value ) {...} 

I do not need to specialize in int, I really need to execute another function for AnotherType. And I cannot change the definition of AnotherType or the base function.

Overloading does not help either because of SFINAE.

Answer 1

You can use BOOST_STRONG_TYPEDEF .

Answer 2

The answer is no. When you enter typedef, you create an alias for the type, not the actual type itself. The compiler will refer to the same thing. That's why:

 typedef int Foo; typedef int Bar; Bar bar = 1; Foo foo = bar; 

Will compile. They are both ints.

Answer 3

I am sure that you cannot use the int compiler and AnotherType in different ways. All typedef — types of aliases — in fact, it does not create a new type; by typedef definition, the compiler will process int and AnotherType equivalently in all cases.

If you need a type with int that is handled differently, you just need to create a one-part struct . Most operations with int content will be compiled with the same machine code as the empty int, but now your data type may have its own specialized templates, etc.

Answer 4

And I cannot change the definition of AnotherType or the base function.

Then you are screwed. I'm sorry. The only option you have, a strong typedef, is not an option if you cannot change the definition to use a strong typedef.

Answer 5

The compiler will treat both specializations the same way, as AnotherType is just a different name for int . You say you don’t need to specialize in int , so just completely remove this specialization and let it specialize in any type of AnotherType that will work.