Re: std::string::npos always < std::string::size() ?

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 10 Jan 2008 02:42:47 -0800 (PST)
Message-ID:
<c04f4075-cec9-49b7-b943-195389dc0800@h11g2000prf.googlegroups.com>
On Jan 10, 4:52 am, "Christopher Pisz" <some...@somewhere.net> wrote:

Is std::string::npos always going to be less than any
std::string 's size()?


It's always going to be greater.

I am trying to handle a replacement of all occurances of a
substr, in which the replacement also contains the substr.
Yick. All I could come up with is:

#include <string>

int main()
{
   std::string text;
   text = "\nThis is a test line with newlines\n<-here and \n<-here and
\n\n<-two here";

   // Format the text
   std::string formattedText(text);

   // Change every occurrance of "\n" to "\r\n"
   std::string::size_type index = 0;

   do
   {
      index = formattedText.find('\n', index);
      if( index != std::string::npos )
      {
         formattedText.insert(index, "\r");
         index += 2;
      }
   } while( index < formattedText.size() );

   // debugging contents of formattedText here
   return 0;
}

However, it depends on std::string::npos always being less
than the size of the string and I am not certain whether that
is safe or not.


Your code should work, although I'd definitely use a while or a
for, rather than a do...while:

    for ( size_t pos = text.find( '\n' ) ;
            pos != std::string::npos ;
            pos = text.find( '\n', pos + 2 ) ) {
        text.insert( pos, '\r' ) ;
    }

But generally, I'd rather not do this sort of thing in place.
Something more like:

    std::string
    convertToDos( std::string const& source )
    {
        std::string result ;
        for ( std::string::const_iterator it = source.begin() ;
                it != source.end() ;
                ++ it ) {
            switch (*it) {
            case '\n' :
                result += "\r\n" ;
                break ;

            default :
                result += *it ;
                break ;
            }
        }
        return result ;
    }

This seems a lot cleaner to me, despite having more lines of
code, and is probably faster as well (less copying). If the
strings are very long, and performance is still a problem, you
can add the following after the definition of result:

    result.reserve( source.size()
                    + std::count( source.begin(), source.end(),
'\n' ) ;

. Unless the strings are very long, however, it probably won't
matter.)

--
James Kanze (GABI Software) mailto:james.kanze@gmail.com
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