Re: [Boost-users] split_iterator<string::iterator> example

From: Pavol Droba (droba_at_[hidden])
Date: 2007-02-20 15:52:47

Hi,

Meryl Silverburgh wrote:
> On 2/20/07, Pavol Droba <droba_at_[hidden]> wrote:
>>> Pavol,
>>>
>>> Thank for your help. I thought you said 'The question marks can be
>>> substituted by
>>> string_split::value_type'. In one of our previous exchange.
>>>
>>> That is why I am using 'string_split::value_type' not
>>> 'iterator_range<string::iterator>'
>>>
>>> Sorry if I mis-understand what you teach me.
>>>
>> No worries.
>>
>> There is no problem with using string_split::value_type. It is just
>> a typedef to iterator_range<string::iterator>. In your code, you
>> misplaced it with string_split itself.
>>
> 
> I change the integer_compare class to use  iterator_range<string::iterator>&.
> 
> And I am trying to loop thru the iterator_range inside the function()
> of integer_compare:
> 
> 	bool operator() (const 	iterator_range<string::iterator>& a,
> 	
> 	const 	iterator_range<string::iterator>& b) {
> 		cout << "calling integer_compare " << endl;
> 		
> 		// expect to print out "", "1", "1", "2" for a string of "/1/1/2"
> 		for (string::iterator sitr = a.begin(); sitr != a.end(); sitr++) {
> 			
> 			cout << copy_range<std::string>(*sitr) << endl;
> 			
> 		}
>                return true;
>        }
> 

Please, see the iterator_range documentation:
http://www.boost.org/libs/range/doc/utility_class.html#iter_range

Although you changed the names of parameters, you are still using 
iterator_range as if it was a split_iterator.

copy_range copies a *range* to a given container. when you dereference
a string::iterator, you will not get a *range*, rather a single character.

Instead of "for" loop you should use just
cout << copy_range<std::string>(a) << endl;

Regards,
Pavol.