Hi, On Fri, Mar 03, 2006 at 02:42:14PM +0100, Olaf van der Spek wrote:
Pavol Droba wrote:
The functionality is in cvs. I have added lexicographical_compare (also in i variant) and comparison predicates is_(i)less, is_not_(i)greater.
Check it out.
struct is_not_greater { template< typename T1, typename T2 > bool operator()( const T1& Arg1, const T2& Arg2 ) const { return Arg1>=Arg2; } };
=, isn't that is_not_less?
Oh, this is what happens when the regression tests are not complete. Thanks for notice, you are right of couse. The operator should be reversed.
struct is_iless { is_iless( const std::locale& Loc=std::locale() ) : m_Loc( Loc ) {}
template< typename T1, typename T2 > bool operator()( const T1& Arg1, const T2& Arg2 ) const { #if defined(__BORLANDC__) && (__BORLANDC__ >= 0x560) && (__BORLANDC__ <= 0x564) && !defined(_USE_OLD_RW_STL) return std::toupper(Arg1)<std::toupper(Arg2); #else return std::toupper(Arg1,m_Loc)<std::toupper(Arg2,m_Loc); #endif } }
What about 'A' vs 'a'? Shouldn't 'A' < 'a'?
Hmm, interesting point. But it is not fully correct. Take this example: str1="Aab" str2="aAa" in my opinion following should hold str1>str2, however if 'A' < 'a' then the comparison would yield the oposite. Such a comparison make sense only if both strings are equal (except the case). Current implementation of ilexicographical_compare does not perform this extra step since it tries to be compliant with std::lexicographical_compare.
I'm also wondering how expensive std::toupper is. Would it make sense to do a == first to avoid the std::toupper calls in certain cases?
Well, this is a more complicated point. Small optimization you are proposing does not really make any difference. I have read a paper (I don't remember where), where author proposes to cache the results of tolower during the comparison. This approch is complex and it effectively disables the possibility to use user-defined comparison predicates. So I decided to go for a simple solution. Best regards, Pavol