Increment operator in c

Are you ready for some serious brain storming?
Here it goes my first post.

1. What would be the output of this program while running with the gnu gcc compiler?


#include<stdio.h>

int main(int argc, char * argv[]){

      int z=2;
      int y=2;
   
      z=z++;
      y=++y;
      printf("%d  %d",z,y);
 return 0;
}

Please Before proceed to answer just think for a minute.
Think ::::


This Space is left intentionally.









Although this is undefined behavior due to absence of sequence point. I will define sequence point in my future posts. But here I am presenting one of the reasons of getting this output.




Answer is  : 2 3

Reason is : For understanding the reason behind this we have to understand the execution sequence of instruction z=z++;

For evaluation of a expression compiler generates a temporary variable and evalute the expression and store the value in the temporary variable which in the given expression is 2 because postfix operator executes after the expression is evaluated. So till now the temporay variable has value 2 

and now z becomes three due to posfix operator. And then the temporary variables value is assigned to the variable which is on the left side of the = operator which here is z itself so value 2 is assigned to z again.

Finally the value of z becomes 2.

Similarly for the value of y first expression is evaluated and in this pre increment is also evaluated so value of y becomes 3 and the value of temporary variable is also becomes 3 and now this temporary variable's value is assigned to y.
Finally the value of y becomes 3.