Re: [Boost-users] [string_algo] split input parameter not const?

From: Pavol Droba (droba_at_[hidden])
Date: 2005-12-13 13:24:49

On Tue, Dec 13, 2005 at 11:01:25AM -0600, Thore Karlsen wrote:
> 
> split() is declared as follows:
> 
> template<typename SequenceSequenceT, typename RangeT, typename
> PredicateT> SequenceSequenceT & 
>   split(SequenceSequenceT & Result, RangeT & Input, PredicateT Pred, 
>         token_compress_mode_type eCompress = token_compress_off);
> 
> Is there a reason why Input isn't const? It would be nice to be able to
> construct an iterator range in the call to split(), like this:
> 
>  typedef iterator_range<string::const_iterator> range;
>  vector<range> inputs;
>  split(inputs, range(Begin, End), is_any_of("\r\n"), token_compress_on);
> 
> It's the same for find_all(), ifind_all(), and probably others, so there
> might be a good reason for it that I'm missing.
> 

The const is not missing there. Because the function is templated,
const is added to the calculated type.

Example:

std::string str("Hello world");
const std::string cstr("Good bye world");

split(res, str, pred); // RangeT is substituted to std::string
split(res, cstr, pred); // RangeT is substituted to const std::string

So you can use the code you have described without problems.

Regards,
Pavol.