Can anyone explain the output process in java

Question

int p=15, q=12, s;  
s=p/q*++p%(++p%--q);
System.out.print(s);

I am trying it according to the precedence of operators and associativity but I think it should give me the output 2 but it is giving me 4 as output.

Could you please remove the parts that you understand and only show the operations that you think should be working differently? — rethab, Nov 25 '21 at 12:48
Does this answer your question? [What are the rules for evaluation order in Java?](https://stackoverflow.com/questions/6800590/what-are-the-rules-for-evaluation-order-in-java) — Hulk, Nov 25 '21 at 12:51
It is essential to understand that order of evaluation is distinct from operator precedence. Order of evaluation is left to right. — Hulk, Nov 25 '21 at 12:52
Also see https://stackoverflow.com/questions/29162301/why-is-the-operator-precedence-not-followed-here?noredirect=1&lq=1 — Hulk, Nov 25 '21 at 12:54

Zabuzard · Answer 1 · 2021-11-25T12:59:22.170

Precedence

p/q*++p%(++p%--q)

is computed as

((p / q) * (++p)) % ((++p) % (--q))

using the operator precendece rules.

Note that ++x/--x increment/decrement the variable by one and then return the result after the operation. Since they change the variable itself, the effect obviously carries over to other usages of the variable.

Evaluation

It is then evaluated left to right, the math is

((15 / 12) * (16)) % ((17) % (11))

which results in 4:

  ((15 / 12) * (16)) % ((17) % (11))
= (1 * 16) % 6
= 16 % 6
= 4

Can anyone explain the output process in java

1 Answers1

Precedence

Evaluation