On Tue, 13 Dec 2005 19:24:49 +0100, Pavol Droba <droba@topmail.sk> 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.