Thread: Getting error while trying to print out a char pointer

Your code exhibits undefined behaviour, as c has not been initialised so does not point at anything. It is necessary to ensure c points at something valid, before dereferencing *c (even accessing the value of *c gives undefined behaviour, let alone assigning a value to *c as you are)

Code:

   char x;
   char *c = &x;

   *c = '9';   /*  this will change x  */

   printf ("%c",*c);