Date: Wed, 10 May 2006 15:30:08 -0700
From: Rush Manbert < rush@manbert.com>
Subject: [Boost-users] Template instantiation problem
To: boost-users@lists.boost.org
Message-ID: < 44626970.4070704@manbert.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

I have a template class that is designed to contain built in types or
std::string types. I always need to be able to cast an instance of the
class as a std::string, and I also need to be able to cast it as the
parameter type with which it was instantiated.

In my class definition, I have defined two cast operators, like so:
(Extraneous stuff omitted)

template<typename T>
class MyDataObject : public MyDataObjectBase
{

        // Cast as std::string
        inline                  operator const std::string &    ()
                                {
                                        ...some code here
                                }

        // Cast as T
        inline                  operator const T &              ()
                                {
                                        ...some code here
                                }
}

The original code was written on a Mac and compiled with Gnu, which did
not complain. even when I did this:

class MyStringObjectClass : public MyDataObject<std::string>
{
}

Now, however, I have moved the code to Windows, and the Visual Studio
compiler complains about the second cast operator when I define
MyStringObjectClass.

I thought that if I could conditionally define the cast as T operator
(or the cast as std::string) I would be okay, but that means testing
against the value of T with the preprocessor. I don't see how to do
that, but I know that some of you folks are really good at bending the
preprocessor to your will.

Can anyone show me a way out of this predicament?

Thanks,
Rush


Take a look at  enable_if.  ie boost::mpl::enable_if I think.  Also is_same.
With those you can check if your T is_same as std::string, and enable (or disable) your operators accordingly.

Tony.