[boost] Re: How to convert a template parameter into a string

From: Dirk Gerrits (dirk_at_[hidden])
Date: 2003-03-01 02:08:14

Robert Allan Schwartz wrote:
>>>Perhaps my spelling class template could be folded into type_traits?
>>
>>[code snipped]
>>
>>Looks interesting, but I'm not sure if it's such a huge advantage over:
>>
>>template <typename T>
>>void foo(T)
>>{
>>     cout << "T is " << typeid(T).name() << endl;
>>}
>>
>>The resulting string of your method is more portable of course, but is
>>that the only reason?
> 
> 
> No. There are other reasons why typeid() is not as "good" as my proposal:
> 
> 1) You must #include <typeinfo> in order to use typeid(). This seems to me
> to be unnecessary overhead.

But wouldn't your method need an #include "spelling.hpp"?

> 2) The following program:
> 
> class base { };
> 
> class derived : public base { };
> 
> int main(void)
> {
>  base * b = new derived;
> 
>  base const * const b2 = new derived;
> 
>  foo(b);
>  foo(*b);
>  foo(b2);
>  foo(*b2);
> 
>  return 0;
> }
> 
> produces:
> 
> T is P4base
> T is 4base
> T is PC4base
> T is 4base
> 
> when compiled by g++ and executed in cygwin.
 >
> As you point out, the string returned by typeid().name() is not specified by
> the Standard, so it is not portable, but in this case, it is extremely
> difficult to decipher.

Ouch. That looks horrible indeed.

> I believe a "standardized" (within Boost), portable, and *readable* text
> representation of T makes my proposal better than typeid().

At what cost? The strings you generate are indeed very readable, but 
specializing spelling for every type that might be a template parameter 
someday seems like too much of a burden to me.

Others might disagree though...

Dirk Gerrits