RE: [Boost-Users] pointer traints ?

From: Mark Storer (mstorer_at_[hidden])
Date: 2002-09-24 17:33:50

Couldn't you do something along the lines of:

template < typename PointerT, typeName RefereceT = &(*PointerT) >

Only I think that's an address rather than a reference...

How about:

template < typename T >
void f( T *ptr )
{
      T &ref = *ptr;
      // ... do something with ref ...
}

And what do you have against "->"?  I don't think a reference's "." is any
faster than a pointer's "->", so that shouldn't be a concern... just trying
to save some knuckle grease?   I don't get it.  (I'm getting a lot of
practice with that phrase since I started digging into STL & Boost)

--Mark Storer
  Software Engineer
  Cardiff Software

#include <disclaimer>
typedef std::disclaimer<Cardiff> Discard;

-----Original Message-----
From: Bohdan [mailto:yg-boost-users_at_[hidden]]
Sent: Monday, September 23, 2002 7:38 AM
To: boost-users_at_[hidden]
Subject: [Boost-Users] pointer traints ?

I have following code:

template < typename PointerT >
void f( PointerT ptr )
{
      ReferenceT ref = *ptr;
      // ... do something with ref ...
}

The problem is that ReferenceT is unknown inside f( ... ).
If i had similar code for InteratorT instead of PointerT i was using
iterator_traits.
AFAIK there is no pointer traits for a raw and smart pointers and i can not
use type_traits
for smart pointers.
So, the question is: "Is there some simple solution for this problem or this
is just
lack of pointer_traits in boost (std?) ?"

regards,
bohdan

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