Difference between expressions of pointers

[ekosix]

Could you explain to me what's the difference between these expressions?

p++;

(*p)++;

*(p++);

(assuming p as a pointer)

Thanks

[tabstop]

The first adds one to the pointer itself (making it point somewhere else). The second adds one to the value pointed to by p. The third moves the pointer, then gets the (new) value pointed to.

[sivanihonjin]

I think so the following example will solve ur doubts

#include <stdio.h>

 main(){
  int *p;
  int c;

  c = 12;
  p = &c;

  printf("%d\n", *p);           // this will print the value which it points
  //  printf("%d\n", p++);    // this wil point to some location                                                                                    
  printf("%d\n", (*p)++);    // this will print the original value which it points and then it increments
  printf("%d\n", *(p++));    // this will print the one time original value and address of p gets incremented
  printf("%d\n", *(p++));    // since address of p gets incremented it will print some value in it 
  return 0;
}
--------------------
the output is 
12
12
13
-1074905056

[MK27]

To clarify:

  printf("%d\n", *(p++));    // this will print the one time original value 
                     and address of p gets incremented
  printf("%d\n", *(p++));    // since address of p gets incremented 
                     it will print some value in it

p++ is post increment, which means after the expression is evaluated (++p is pre increment). So the first time p++ is called, p's address increases by the unit size of it's type, in this case an int. In the case of sivanihonjin's code, this is now out of bounds so this should not have been done. Subsequently accessing p's value may cause a segmentation fault. If not, you will get whatever value happens to be where p is pointing to.

[sivanihonjin]

ThanKs for your suggestion:
Yes its true.
p's value may cause a segmentation fault. If not, you will get whatever value happens to be where p is pointing to.

[quzah]

To be pedantic, it's not the value that causes the segfault. It's the dereferencing of the pointer that causes it. (i.e.: trying to access something you shouldn't)


Quzah.