Re: Class template specialization with template parameter

Victor Bazarov <>
Wed, 14 May 2008 18:28:19 -0400
Greg Herlihy wrote:

On May 14, 12:51 pm, flopbucket <> wrote:


First, thanks for the reply.

Foo<int> uses the normal template, but Foo<std::map<T1, T2> > for any
types of T1 and T2 uses the specifalization?

Not sure what you're asking here.

What I was trying to say was that if I declare:

Foo<int> myFoo();

I believe that you mean to declare an object (and not a function):

   Foo<int> myFoo;

The instantiation will be for the normal/not-specialized template.


But if I declare:

Foo<std::map<int, std::string> > myFoo2();

Again, I believe that you want:

   Foo<std::map<int, std::string> > myFoo2;

it will use the specialization... and that regardless of the types
used for std::map (in this case, int and std::string), any
"Foo<std::map<typeA, typeB> > myFooExample()" will always
instantiate the specialization.

The answer depends on how many template parameters "std::map"
requires. The usual number is two - the two specified by the C++
Standard. An implementation however is allowed to require additional
temmplate type parameters for a std::map. So, assuming that std::map
requires only two type parameters, then the answer is "yes".

I don't think it has anything to do with the number of arguments the
template takes/has. The specialisation is not a template with a
template template argument (I believe that's what you're thinking of...)

For instance,

     template<class T, class U = int> class TwoArgs {};
     template<class T> struct Foo { enum {a=0}; };
     template<class T> struct Foo<TwoArgs<T> > { enum {a=1}; };

     int main() {
         char should_complain[Foo<char>::a];
         char dont_know[Foo<TwoArgs<int,char> >::a];
         char should_be_OK[Foo<TwoArgs<double> >::a];

The declaration of 'should_complain' instantiates the regular 'Foo'
because it's not the specialisation on 'TwoArgs'. The declaration of
'should_be_OK' is the specialisation which has 'a == 1'. Now, since
'dont_know' uses the 'Foo' instantiated for 'TwoArgs', yet the second
argument of 'TwoArgs' is not the default (int), I am not sure. The
Comeau online test drive does not like 'dont_know', most likely because
it uses the 'TwoArgs' in a way different from the specialisation's, and
as the result 'a' is 0. Tricky...

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