Re: std::string::npos always < std::string::size() ?
On Jan 10, 1:29 pm, Abhishek Padmanabh <abhishek.padman...@gmail.com>
wrote:
On Jan 10, 8:52 am, "Christopher Pisz" <some...@somewhere.net> wrote:
Is std::string::npos always going to be less than any std::string 's siz=
e()?
I am trying to handle a replacement of all occurances of a substr, in wh=
ich
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<-her=
e 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.
It is ok/safe. An std::string cannot have a length() (since it is
std::string, size()) greater than std::string::npos. So, your code
should be perfectly fine, it will always exit the do-while when all
instances of '\n' have been dealt with.- Hide quoted text -
- Show quoted text -
Ideally, you should handle the length_error exception.
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