Re: [boost] Re: [Formal review] Collection Traits library

From: Pavol Droba (droba_at_[hidden])
Date: 2004-05-08 13:17:06

On Sat, May 08, 2004 at 10:07:30PM +1000, Thorsten Ottosen wrote:
> | > Pavol uses it havily in his string library. The "dangerous" thing would be not to put
> | > computation of the end-iterator out-side the loop, eg.
> | >
> | > for( iterator_type_of<T>::type i = begin( c ); i != end( c ); ... )
> | >                                                                   ^^^^^^^^^^  ^
> | > is not good. But one wouldn't do that anyway. And if one did it, what are the chances of
> | > the argument being a char*?
> | 
> | Too many chances are being taken here, IMO. 
> 
> the more chances that are taken, the less the likelihood of it all happening.
> 
> | It doesn't feel right.
> | STL's end is guaranteed O(1) and I'm really concerned about anything
> | that violates that expectation.
> 
> ok. I have forgotte why Pavol want to support char* at all. I mean, programming with manual
> insertion of null is really error-prone. If char* is considered a range, one might even say
> 
> std::string s;
> my_algo( s.buffer() ); // oops, no terminating null!
>  

There is a lot of legacy code out there with no counterpart in c++. I don't think, that we can afford 
not to support it. Just have a look at the Boost Regex interface. char* is still there, and 
there are reasons for it.

An implication of supporting char* is the requirement, that char[] must be threated the same way.
If you consider the way how std::string's end() works you have the current specification of the
collection traits for string.

An example of usecase for all string variants:

#include <iostream>
#include <string>
#include <boost/algorithm/string.hpp>

using namespace std;
using namespace boost;

int main(int argc, char* argv[])
{
   // check if the first argument contains "hello"
   // char* is used for input char[] for search specifier
   iterator_range<char*> r=find_first(argv[0], "hello");

   if( !r.empty() )
   {
      // convert the find result (a pair of char*) to std::string 
      string str(r.begin(), r.end());
      str+=" ";
      str+=str;

      cout << str << endl;   
   }

   return 0;
}

An example may not be very usable, but it shows a common pattern, when char*, char[], and std::string
are used in one place. 

Regards,

Pavol