Thread: Simple sscanf mystery

Hello all,
Okay I'm stumped. I must be misreading something, but I think the below program should print "1" as the result of sscanf. That is, I believe searching for "t%c" would catch the character "h" in the below example, but I can't figure out why it always returns 0. Any ideas?

Code:

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

int main(int argc, char *argv[], char *envp[])
{
   /* we only care if sscanf found our substring, we don't care what it's value is */
   char *junk;

   junk = (char *) malloc (25);  /* allocate our memory */

   /* why doesn't this return 1? t%c is within "Hello there", should read 'h' */
   printf("\nsscanf returned: %i\n", sscanf("Hello there", "t%c", junk));

   free (junk); /* free up our memory */

   return (0); /* always return success */

} /* end main(...) */

I hate just flat-out asking for the answer but it seems I fundamentally misunderstand either the syntax or the sscanf function itself. Thanks for any hints or pointers.

Josh

Thanks!

Just one more quick question. Why does junk[2] remain -1 in the below code? Shouldn't junk[2] contain '3456'?

Code:

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

int main(int argc, char *argv[], char *envp[])
{
   int *junk;
   char *data;
   int  i, value;

   data = (char *) malloc (15);
   junk = (int *) malloc (15);

   for (i = 0; i < 15; i++)
   {
      junk[i] = -1;
   }

   strcpy(data, "09+234.88\r\n A151 L+3000 C3456\r\n");

   value = sscanf(data, "%*[^ ] A%i L+%i C%i\r\n", &junk[0], &junk[1], &junk[2]);

   for (i = 0; i < 5; i++)
   {
      printf("\njunk[%i]: %i", i, junk[i]);
   }

   printf("\nvalue: %i\n\n", value);

   free (data);
//   free (junk); SIGSEV :(

   return (0); /* always return success */

} /* end main(...) */

My output is:

junk[0]: 151
junk[1]: 3000
junk[2]: -1
junk[3]: -1
junk[4]: -1
value: 2