[SOLVED] strstr doesn't work with the delimiter "\r\n\r\n"

strstr doesn't work with the delimiter "\r\n\r\n"

i got the delimiter "\r\n\r\n" in the substring, and strstr is returning null

Here is the code :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


int main(int ac, char **av) {
  char *ptr;

  ptr = strstr(av[1], "\r\n\r\n");
  printf("ptr = %s\n", ptr);
  return 0;
}

I launch the code with this :

./a.out "POST /cgi-bin/process.cgi HTTP/1.1\nUser-Agent: Mozilla/4.0 (compatible; MSIE5.01; Windows NT)\nHost: www.tutorialspoint.com\nContent-Type: text/xml; charset=utf-8\nContent-Length: length\nAccept-Language: en-us\nAccept-Encoding: gzip, deflate\nConnection: Keep-Alive\r\n\r\n<?xml version='1.0' encoding='utf-8'?>"

And ptr is equal to (null), why ?

Solution

The string you passed to the code contains the eight character sequence \, r, \, n, \, r, \, n.

The string literal "\r\n\r\n" produces the four character sequence ␍, ␊, ␍, ␊.

To produce a string that would match the argument, use the following string literal:

"\\r\\n\\r\\n"

But I think it's more likely you want to providing a proper HTTP request.

Depending on which echo you use, echo or echo -e might produce the desired string (plus a trailing line feed):

$ echo -e 'a\r\nb\r\n' | od -c
0000000   a  \r  \n   b  \r  \n  \n
0000007

printf can reliably produce exactly the string you want, though you have to escape % symbols by duplicating them.

$ printf 'a\r\n%%\r\n' | od -c
0000000   a  \r  \n   %  \r  \n
0000006

Example usage:

./a.out "$( printf 'POST ...\r\n\r\n...' )"