difference between *y++ and ++*y?

Question

I'm confused in how this code will get executed. Suppose we have

int x=30,*y,*z;
y=&x;

what is the difference between *y++ and ++*y? and also what will be the output of this program?

#include<stdio.h>
int main(){

    int x=30,*y,*z;
    y=&x;
    z=y;
    *y++=*z++;
   x++;
   printf("%d %d %d ",x,y,z);
   return 0;
}

Answer 1

The expression x = *y++ is in effects same as:

x = *y;
y = y + 1;

And if expression is just *y++; (without assignment) then its nothing but same as y++;, that is y start pointing to next location after increment.

Second expression ++*y means to increment the value pointed by y that same as: *y = *y + 1; (pointer not incremented) It will be better clear with answer to your first question:

Suppose your code is:

int x = 30, *y;
int temp;
y = &x;

temp = *y++; //this is same as:  temp = *y;  y = y + 1;

First *y will be assigned to temp variable; hence temp assigned 30, then value of y increments by one and it start point to next location after location of x (where really no variable is present).

Next case: Suppose your code is:

int x = 30, *y;
int temp;
y = &x;

temp = ++*y;  //this is same as *y = *y + 1; temp = *y;

First value of *y increments from 30 to 31 and then 31 is assigned to temp (note: x is now 31).

next part of your question (read comments):

int x = 30, *y, *z;

y = &x;       // y ---> x , y points to x
z = y;        // z ---> x , z points to x
*y++ = *z++;  // *y = *z, y++, z++ , that is 
              // x = x, y++, z++
x++;          // increment x to 31