Date: Wed, 10 May 2006 15:30:08 -0700

From: Rush Manbert 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 MyDataObjectstd::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.