Re: Class template specialization with template parameter

James Kanze <>
Thu, 15 May 2008 01:32:39 -0700 (PDT)
On May 14, 10:20 pm, Greg Herlihy <> wrote:

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

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

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.

The usual number is four, since that's what the C++ standard
requires. He has defined a partial specialization for Foo<
std::map< T, U, std::less<T>, std::allocator< std::pair< T
const, U > >. If he instantiates Foo with a type which
corresponds to this, the partial specialization will be used.
Otherwise, the non-specialized version will be used. Thus, with
std::map< int, std::string > (or std::map< int, std::string,
std::less< int > >), he'll get the partial specialization, but
with std::map< std::string, int, CaseInsensitiveCompare >, he'll
get the non-specialized version.

An implementation however is allowed to require additional
temmplate type parameters for a std::map.

But only if they follow the required four, and have default
values. So as long as you don't use them, his code should work.

So, assuming that std::map requires only two type parameters,
then the answer is "yes".

Assuming that the implementation of std::map is conform, his
partial specialization will be used for all std::map
instantiations which use std::less<Key> as the comparator, and
std::allocator< std::pair< Key const, T > > as the allocator.
(Using something other than the default allocator is fairly
rare, but it's quite frequent to find map's with other than the
default comparison function. In his case, I would definitly add
that to the arguments of the partial specialization. And it's
not that difficult to add the allocator either:

    template< typename Key, typename Value,
              typename Cmp, typename Alloc >
    class Foo< std::map< Key, Value, Cmp, Alloc > >
        // ...
    } ;

This should work for all instanciations of std::map.

James Kanze (GABI Software)
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