Can one explain this?

Question

int x = 0;
int z = x++ + x++;

Expected behaivour: x = 2, z = 0.
Compiled: x = 2, z = 1.

Why was z incremented when x was both postfix?

Answer 1

The first postfix affects the second one.
It might look like it should be 0, but in fact the first postfix changes the value of x to 1, so the second time you call it, x++ will be 1 and not 0.

EDIT:
A better way to see this is by using variables.
int x = 0;
int z1 = x++; // so z1 = 0, and x = 1
int z2 = x++ // since x = 1, then x++ returns 1, and x becomes 2
int z = z1 + z2 = 0 + 1 = 1

Answer 2

Prathvi you explained the wrong output..

Answer 3

Aymane Boukrouh Thank you, I think I get it now. Always thought that postfix will increment only after the statement, no matter what.

Answer 4

x++ return the value of x than increment means
int x=0
int z= 0/* x=1*/+ 1/*x=2*/;
So x= 2 and z=1

Like
x=0
z=x++;//z=0 and x=1