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

From: Thore Karlsen (sid_at_[hidden])
Date: 2005-12-13 14:00:05

On Tue, 13 Dec 2005 19:24:49 +0100, Pavol Droba <droba_at_[hidden]>
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.

Hmm.. I can use a const value just fine, but I can't use a temporary. At
least not in VC++ 8.0 with the warning level set to 4 (I compile all my
code at warning level 4). I get a warning C4239:

warning C4239: nonstandard extension used : 'argument' : conversion from
'boost::iterator_range<IteratorT>' to 'range &'
 with
 [

IteratorT=std::_String_const_iterator<char,std::char_traits<char>,std::allocator<char>>
 ]
 A non-const reference may only be bound to an lvalue

-- 
Be seeing you.