Re: Template question

From:
Gert-Jan de Vos <gert-jan.de.vos@onsneteindhoven.nl>
Newsgroups:
comp.lang.c++
Date:
Wed, 2 Dec 2009 08:15:49 -0800 (PST)
Message-ID:
<1b4042bd-a91b-4749-8aec-2e80bdbf392c@r24g2000yqd.googlegroups.com>
On Dec 2, 5:00 pm, Adrian <n...@bluedreamer.com> wrote:

Hi All,

Is the output from the following code correct? Is the result
undefined? I am not quite sure what I expected from this :-). Should
both the time_t and DateTime creation call the specialized class or
should only the time_t. Both results would make sense to me.

Does it say this anywhere in the standard at all?

TIA

Adrian

------------------------------------
#include <iostream>
#include <ctime>

typedef time_t DateTime;

class BoringBase
{
   public:
      virtual ~BoringBase() {}

};

template<typename T>
class Foo : public BoringBase
{
   public:
      Foo() { std::cout << "Constructing run-of-the-mill Foo\n"; }

};

template<>
class Foo<time_t> : public BoringBase
{
   public:
      Foo() { std::cout << "Special time_t Foo\n"; }

};

int main(int /* argc */, char /* *argv[] */)
{
   BoringBase *a,*b,*c;

   a=new Foo<int>();
   b=new Foo<time_t>();
   c=new Foo<DateTime>();

   return 0;

}

// Outputs:
// [adrianc@mlx32dev:~]$g++ -Wall -ansi -pedantic -Wextra -Weffc++
foo.cc
// [adrianc@mlx32dev:~]$a.out
// Constructing run-of-the-mill Foo
// Special time_t Foo
// Special time_t Foo


This result is correct. The typedef does not create a new type, it
defines a synonym for time_t named DateTime. Both are names for one
and the same type, therefore both b and c show the same behavior. Use
struct, class, union or enum to define new types.

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