page 99 OCA 8 Guide: Review Question 19: post-unary decrement

[Pedro Esgueira]

Hi,

I thought I was understanding post-unary increment/decrements, but apparently not.

code:

int m = 9, n = 1, x = 0;
while(m > n) {
    m--;
    n += 2;
    x += m + n;
}
System.out.println(x);

My doubt:
I'd understood from the explanation on the book on page 58 that "if the operator is placed after the operand, [...], then the original value of the expression is returned, with operator applied after the value is returned." The examples of page 59 consubstantiate this. Hence, I thought that in the first iteration while (9 > 1)  m would be 9, and 8 or 7 in the second or third loop, respectively.
But I'm wrong, and 've compiled and checked for each loop the value of m to be 8, 7 and 6, respectively.

What am I not grasping about post-unary operators?

[Campbell Ritchie]

Please call x++ and x-- postfix operators, not post‑unary.
By the time you reach line 4, the new value of m is visible. That is 8, 7, 6.

[Pedro Esgueira]

Hi Campbell,

Campbell Ritchie wrote:By the time you reach line 4, the new value of m is visible. That is 8, 7, 6.

Ok, I see, so the post-increment operator becomes visible because the next line is a new expression (so: it's per expression, not per loop, that I need to think about this operator). Thanks

[Campbell Ritchie]

Yes. The new value of the variable incremented by x++ is visible immediately after that expression. But code using more than one ++ or -- operator per statement is bad code which should only be seen in cert exams