Problem when using the strchr function.

Linh · 05-07-2004, 05:36 PM

Problem when using the strchr function.
The code below result in a null value despite the fact that
a colon is in the string. This is incorrect
ip_address = strchr(ip_address_string, ':');

The code below result in an 'r' value because the letter
r is in the string. This is correct
ip_address = strchr(ip_address_string, 'r');

===============================
root:/home# ./string_1
ip_address = (null)
===============================

Code:

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

main ()

{
  char ip_address_string[14] = "addr:10.4.0.1";
  char *ip_address;

  ip_address = strchr(ip_address_string, ':');
  printf ("ip_address = %s\n", ip_address);

}

itsme86 · 05-07-2004, 05:47 PM

I don't know...it works okay for me:

Code:

itsme:~/C$ cat strchr.c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

main ()
{
  char ip_address_string[14] = "addr:10.4.0.1";
  char *ip_address;

  ip_address = strchr(ip_address_string, ':');
  printf ("ip_address = %s\n", ip_address);

}
itsme:~/C$ ./strchr
ip_address = :10.4.0.1
itsme:~/C$

Linh · 05-07-2004, 06:02 PM

Hi itsme86. Thank you for your help.
I used the function s = strtok(ip_address_string, ":")
and then I used the function
ip_address = strchr(ip_address_string, ':') and that is why
it gives me a null value.

The function strtok kept a hidden static pointer.
When I called the strchr function afterward, the pointer points to null.
If I commented out the code s = strtok(ip_address_string, ":")
it gives a correct answer when I run the code

ip_address = strchr(ip_address_string, ':')

===================================

Code:

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

main ()

{
  char ip_address_string[14] = "addr:10.4.0.1";

  char *s;
  char *save_pointer;
  char *ip_address;


  s = strtok_r(ip_address_string, "addr:",  &save_pointer);
  printf ("using strtok_r function.  s = %s\n", s);


  s = strtok(ip_address_string, ":");
  printf ("using strtok   function.  s = %s\n", s); 


  ip_address = strchr(ip_address_string, ':');
  /* ip_address = (strchr(ip_address_string_2, '1') + 1); */
  printf ("ip_address = %s\n", ip_address);

}

pblack15 · 08-21-2006, 08:46 AM

I am having a similar problem with strchr with SUSE Linux 10.1. The following code compiles and works fine on MS-Dos and SCO Unix 5.06 but generates a Segmentation fault on SUSE Linux 10.1.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
main()
{
static char buffer[80];
void converter();

fprintf(stdout,"Try entering 90-45-9\n");
fgets(buffer,79,stdin);
converter(buffer);
fprintf(stdout,"Try entering 90.775\n");
fgets(buffer,79,stdin);
converter(buffer);
return 0;
}
void converter(char *buffer)
{
char *p, *q;
double d, e, f;

d = atof(buffer);
p = strchr(buffer,'-');
p++;
if(*p == '-')
{
e = atof(p);
q = strchr(p,'-');
if(*q == '-')
{
q++;
f = atof(q);
}
}
fprintf(stdout,"%f %f %f\n",d, e, f);
}

xhi · 08-21-2006, 09:05 AM

welcome to linux questions. in the future please use code tags as they preserve the formatting and make the code alot easier to read.

as for your segfault, it might have something to do with you asking for a .

> fprintf(stdout,"Try entering 90.775\n");

and then looking for a -

> strchr(buffer,'-');

then you continue through converter() without verifying that p is valid. if you were to add this block after p = strchr(buffer, '-'); in converter.. you should see where you are going wrong.

Code:

if(!p)
{
    printf("did not find char in string, bailing out\n");
    return;
}

pblack15 · 08-22-2006, 08:36 AM

Follow up on my earlier post:

Use p = strchr(buffer,'-');
if(p == NULL)

NOT p = strchr(buffer,'-');
if(*p == NULL)

xhi · 08-22-2006, 10:39 AM

Quote:

Originally Posted by pblack15

Follow up on my earlier post:

Use p = strchr(buffer,'-');
if(p == NULL)

NOT p = strchr(buffer,'-');
if(*p == NULL)

im sorry maybe its just because i cant read minds very well, but i have not idea what you are talking about.. care to elaborate?