I think I either found an error in the OCA SE 8 Study Guide book for just don't understand it.
In Ch 2 pg 59 there is the following example snippet:
The explanation of the result is as follows.
First, the x is incremented and returned to the expression, which is multiplied by 5. We can simplify this:
Next, x is decremented, but the original value of 4 is used in the expression, leading to
The final assignment of x reduces the value to 2, and since this is a pre-increment operator, that value is returned to the expression:
Finally, we evaluate the multiple and division from left-to-right ....
The explanation does arrive at the correct values for x and y but, I think the order of ops is described incorrect. Shouldn't it be that the Post-unary operators are evaluated first (x--) as described in "Table 2.1 Order of operator precedence" on pages 52-53 which in this case is as follows below?
It is not simply a case of precedences, but the fact that all operations are evaluated left to right. The leftmost operator (preincrement) is applied to x first. The fact that postincrement has a higher precedence does not mean that x-- after the divide sign is evaluated first. The left to right rule takes precedence. The JLS (Java® Language Specification) tells you that the left operand is evaluated first, so the x-- token will have the value 4.
Please explain what your line 5 means; I think it is mistaken.
No. Operator precedence does not determine order of evaluation. Operator precedence only determines which operand goes with which operator. You can imagine it as the compiler inserting explicit parentheses in the expression:
After it's done this, operator precedence and associativity no longer play a role. Evaluation is ALWAYS left-to-right.
Consider the expression -x++. The higher precedence of unary postfix operators as compared to unary prefix operators ensures that this expression is interpreted as -(x++), and not as (-x)++.
Roses are red, violets are blue. Some poems rhyme and some don't. And some poems are a tiny ad.