[boost] Re: notation question

From: Vladimir Prus (ghost_at_[hidden])
Date: 2003-01-08 11:41:25

David Abrahams wrote:
> Vladimir Prus <ghost_at_[hidden]> writes:

>>I'd prefer the latter variant, so that non-broken platforms use more natural
>>syntax. Another question is whether we could use only the second version.
>>Theoretically, if you call the second version on const string, then InputT
>>should be deduced as "const string". You can then write a simple wrapper to
>>select const_iterator or iterator.
>>
>>The only problem is, IIRC, borland drops "const" on template arguments
>>sometimes, and that's not possible to fix. Also, the "simple wrapper"
>>requires partial specialization. So, I'm not sure this approach is
>>viable, either.
> 
> 
> There are mostly-transparent solutions.  Something like:
> 
>     // find_first sequence const version
>     template< typename InputT, typename SearchT >
>     inline iterator_range< typename InputT::const_iterator >
>     find_first_impl( const InputT& Input, const SearchT& Search, 0 )
>     {
> 		...
>     }

I'm guessing the last parameter should be "int"?

>     // find_first sequence non-const version
>     template< typename InputT, typename SearchT >
>     inline iterator_range< typename InputT::const_iterator >
>     find_first_impl( InputT& Input, const SearchT& Search, int )
>     {
> 		...
>     }
> 
>     // find_first sequence
>     template< typename InputT, typename SearchT >
>     inline iterator_range< typename InputT::iterator >
>     find_first( InputT& Input, const SearchT& Search )
>     {
> 	find_first_impl(Input, Search, 0);
>     }
> 
> Should work on vc6.  Can't vouch for Borland, though :(

Unfortunately, "first_first" will return the same plain (non-const) iterator
in all cases. And we'd need to return const_iterator when string is const.

- Volodya