Does promotion occur before increment in Java?

Question

I've seen this question in OCA questions and need to know why it outputs 90 and not 100. Here is the code:

int x = 9;
long y = x * (long) (++x);
System.out.println(y);

So, what I think this would do is, firstly, increment x (because that's what happens at first right?) and then it would do the type promotion and take left x which is 10, turn it into long and multiply those two longs. Right?

@Guy, he said what it prints, he wants to know why it prints what it does. — M. Prokhorov, Mar 13 '17 at 12:13
I'm not sure what the casting to long is contributing to this scenario. You get the same thing with `int y = x * (++x)` — khelwood, Mar 13 '17 at 12:15
@khelwood Next question could be "What if x = Integer.MAX_VALUE ?". But that's only speculation on my side. — Fildor, Mar 13 '17 at 12:17
@Fildor Wait what? No, dude. I asked this question because I thought it had something to do with type promotion. That's all. — user218046, Mar 13 '17 at 12:18
No problem. It's from a textbook or something, right? @khelwood Hm, you're right ... — Fildor, Mar 13 '17 at 12:19

score 9 · Accepted Answer · answered Mar 13 '17 at 12:13

9

No. The operands of each operator are evaluated from left to right. Therefore the first operand of the * operator, x, is evaluated before the second operand (long) (++x). Therefore 9 is multiplied by 10.

answered Mar 13 '17 at 12:13

Eran

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An example of where precedence doesn't match order of evaluation. +1 – Peter Lawrey Mar 13 '17 at 12:49
@Eran, So basically, if Java sees variable in operation - it tries to get its values and write in operation right? from left to right. O.K. I understand now. Thank you Sir. – user218046 Mar 13 '17 at 12:49

Does promotion occur before increment in Java?

1 Answers1