Boost Users :
From: Gottlob Frege (gottlobfrege_at_[hidden])
Date: 2006-05-10 23:36:29
Date: Wed, 10 May 2006 15:30:08 -0700
> From: Rush Manbert <rush_at_[hidden]>
> Subject: [Boost-users] Template instantiation problem
> To: boost-users_at_[hidden]
> Message-ID: <44626970.4070704_at_[hidden]>
> 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
> 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?
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.
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