Re: template specialization by another templated type

From:
=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 30 Jun 2009 12:38:57 CST
Message-ID:
<cacc349f-83fc-4c48-adf6-1bda761ab21a@d32g2000yqh.googlegroups.com>
On 29 Jun., 23:29, Aston Martin <masoom.sha...@gmail.com> wrote:

I am trying to specialize a templatized method with another type which
itself takes template parameters.

say e.g i have

template<typename T> void foo() {}

this can be used as foo<int>() or foo<std::vector<int> >, right ?
right.
then consider I want to do something 'special' only when T happens to
be std::vector<T> which itself is template parameterized. how do i do
that without changing the method names ? can i specialize foo<>() on
std::vector<T>


Not directly so, but it is possible to do that in an indirect way,
see below.

is something like below legal ?
template<typename T> void foo<std::vector<T> >() {}


No, because this would be an invalid attempt to
define a partial template specialization for a function
template.

consider the code below
*********************************** START
**********************************
#include <iostream>
#include <complex>

class Junk
{
public:
     template<typename T>
     void fooMeth()
     {
         // so something for T, straight forrward
         std::cout << "fooMeth<" << typeid(T).name() << ">()" <<
std::endl;
     }

     template<typename T>
     void fooMeth()
     {
         // do something specific to std::complex but common to all
std::complex<T>
         std::cout << "fooMeth<std::complex<" << typeid(T).name() << ">

()" << std::endl;

     }
};

int main( int argc, char* argv[])
{
     Junk j;
     j.fooMeth<int>();
     j.fooMeth<std::complex<int> >();
     return 0;}

*********************************** END
**********************************

while i try to compile above code i get this error. I wonder if am i
wrong ?


Yes, because you violated the one-definition-rule
in class Junk by trying to provide two definitions
of the same member function fooMeth() in the same
translation unit.

is my compiler under powered ? or is it language short
coming ?


Neither of.

To realize what you want to do you could take advantage
of partial specialization of class templates, e.g.

#include <typeinfo>
#include <iostream>
#include <complex>

template<typename T>
struct FooImpl {
   static void call() {
     std::cout << "fooMeth<" << typeid(T).name() <<
       ">()" << std::endl;
   }
};

template<typename T>
struct FooImpl<std::complex<T> > {
   static void call() {
      // do something specific to std::complex but common
      // to all std::complex<T>
      std::cout << "fooMeth<std::complex<" <<
        typeid(T).name() << ">>()" << std::endl;
   }
};

class Junk
{
public:
      template<typename T>
      void fooMeth()
      { FooImpl<T>::call(); }
};

int main()
{
      Junk j;
      j.fooMeth<int>();
      j.fooMeth<std::complex<int> >();
      return 0;
}

HTH && Greetings from Bremen,

Daniel Kr?gler

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