Template Specialization for a function without Parameters

Question

I need to specialize a function template in c++.

template<typename T>  
void doStuff<T>() {}

To

template<>
void doStuff<DefinedClass>();

and

template<>
void doStuff<DefinedClass2>();

I guess that is not the correct syntax (since it is not compiling). How should I do it?
Also, Since I will have not undefined template parameters in doStuff<DefinedClass>, would it be possible to declare the body in a .cpp?

Note: doStuff will use T wihtin its body to declare a variable.

Solution 1

The primary template doesn't get a second pair of template arguments. Just this:

template <typename T> void doStuff() {}
//                        ^^^^^^^^^

Only the specializations have both a template <> at the front and a <...> after the name, e.g.:

template <> void doStuff<int>() { }

Solution 2

The correct syntax for the primary template is:

template <typename T>
void doStuff() {}

To define a specialisation, do this:

template <>
void doStuff<DefinedClass>() { /* function body here */ }

Solution 3

I guess that is not the correct syntax (since it is not compiling). How should I do it? doStuff will use T wihtin its body to declare a variable.

template<typename T>  
void doStuff() 
{
  T t = T();   // declare a T type variable

}

would it be possible to declare the body in a .cpp?

C++ only supports inclusive mode only, you can't compile separately then link later.

From comment, if you want to specialize for int type:

template<>
void doStuff<int>()
{
}

Comments

[Kerrek SB]

When you say "it is not compiling", you should include the relevant error message.

[Christian Rau]

Does template<typename T> void doStuff<T>() {} even compile in the first place (I think this <T> is invalid).

[Mario Corchero]

Thanks, but still not working. If I write the specialization like: "template<> void <int>doStuff();" I got: "invalid explicit specialization before '>' token"

[Mario Corchero]

If I put" template<typedef T>" I get "template-id `buildBom<int>' in declaration of primary template"

[Synxis]

No, it should be template<> void doStuff<int>(); for specialization and template<typename T> void doStuff(); for general case...

[Kerrek SB]

@Synxis: Thanks! I added that example, and I removed the ... from the first set, because you can't partially specialize functions anyway. Sometimes a less general answer is more helpful :-)