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

From:
"Jim Langston" <tazmaster@rocketmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 10 Jan 2008 01:47:13 -0800
Message-ID:
<_Qlhj.1792$dP7.797@newsfe07.lga>
Christopher Pisz wrote:

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


Actually, the other way around. std::string::npos is always going to be
greater than sny std::string's size.

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 is perfectly fine, however, your while statment can simply be
changed to:

} while ( index != std::string::npos );

And you could even simplify it a bit by rearranging it.

index = 0;
while ( ( index = formattedText.find('n', index ) != std::string::npos )
{
        formattedText.insert(index, "\r");
        index += 2;
}

Now you don't need the if statmeent and to me it is cleaner code.
--
Jim Langston
tazmaster@rocketmail.com

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