Re: Converting a substring to Integer

From:
Reetesh Mukul <reetesh.mukul@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 6 Feb 2008 07:27:45 -0800 (PST)
Message-ID:
<b72ad404-550d-44a0-8e40-3a873e808098@s19g2000prg.googlegroups.com>
On Feb 6, 6:35 am, "Daniel T." <danie...@earthlink.net> wrote:

Reetesh Mukul <reetesh.mu...@gmail.com> wrote:

You can try this:-

#include <iostream>
#include <sstream>
#include <cctype>

int main()
{
  std::string s = "dsdhjsahdk dsdjdsaj 36782367 sdjdhak";

  int j = 0;
  std::stringstream cstr;
  cstr << s;
  int i = 0;
  char ch;

  while( cstr >> ch )
    {
      if( std::isdigit(ch) )
   {
     cstr.putback(ch);
     break;
   }


From here:

      if( ( j = s.find(" ",j) )== std::string::npos )
   {
     cstr.seekg(-1);
     break;
   }

      cstr.seekg(++j, std::ios::beg);


to here. Is this block of code necessary?

    }

  if(cstr)
    {
      cstr >> i;
    }
  std::cout << i;

  return 0;
}


Wouldn't this be a simpler way of doing the same thing?

int main()
{
   std::string s = "dsdhjsahdk dsdjdsaj 36782367 sdjdhak";

   int i = 0;
   std::stringstream cstr( s );
   char ch = 0;
   while( cstr >> ch && !isdigit( ch ) )
   { }
   cstr.putback( ch );
   cstr >> i;
   cout << i;

}


Actually, I assumed that numeric-digits will appear surrounded by
space. This means some thing like {alphabet}*space {digits}+
space({alphabet}*space)*. Your code will extract digits in the
following conditions also:- djshdjshj14524 jdhdss

Regards,
Reetesh Mukul

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