Priority operators in Java

Question

For the next code, what it's z? (Java)

int x = 5;
int y = 10;
int z =++x*y--;

The order of priority is: y--, ++x, *, =. ( https://introcs.cs.princeton.edu/java/11precedence/ ) Why after run the code, z = 60 ?

`++x` evaluates to 6. `y--` evaluates to 10. 6 * 10 is 60. If that's not what you expect, explain why. — khelwood, Jul 20 '18 at 13:40
y-- is post decrement which mean it is decremented after taking the value so y will be 10. ++x is pre increment so it is incremented first and taken value so it becomes 6. so 10 times 6 = 60 — pvpkiran, Jul 20 '18 at 13:40
`int z =++x*y--;` is equal to `int z = (x+1)*y;y=y-1;` that's why — flyingfox, Jul 20 '18 at 13:41
Why is this valid question so downvoted? I really don't understand it's trend. — Nikolas Charalambidis, Jul 20 '18 at 13:42
Ok.. But what is the order for all operations in Java ? Do you see the link? — Mihaela Mihai, Jul 20 '18 at 13:43
@Nikolas OP could really have explain why expect another result than 60. In details, not by saying that y-- should be evaluated first or something, like that would matter. — kumesana, Jul 20 '18 at 13:43
Why do you care about the order of operations? They have no effect in this example. — kumesana, Jul 20 '18 at 13:44
@Lino that page says "In Java, subexpressions are evaluated from left to right". — Andy Turner, Jul 20 '18 at 13:46
@Lino no... It's just that -- operator is applied to y rather than to the multiplication of x and y. That's operator precedence, not order of operations. Operations are evaluated left to right, and that doesn't matter here anyway. — kumesana, Jul 20 '18 at 13:46
@kumesana the order of execution matters here. `++x` means `x=x+1` **before** using x. `y--` means `y=y-1` **after** using `y`. — Luiggi Mendoza, Jul 20 '18 at 13:47
@LuiggiMendoza that's not order of execution, that's meaning of the operator. No matter when you execute ++x, it will give out the same result anyway in this example. — kumesana, Jul 20 '18 at 13:48
@kumesana it's `++x`, not `x++`, and that **matters** for this case... — Luiggi Mendoza, Jul 20 '18 at 13:50
@LuiggiMendoza I meant about the fact that the order of operation doesn't matter in this example, not about whether the final result will be 60 or something else. — kumesana, Jul 20 '18 at 13:52
@kumesana but the order of operation matters because that acts directly on the output... — Luiggi Mendoza, Jul 20 '18 at 13:52
@LuiggiMendoza what output? Here the only output we have is the computed value of the expression. Order of operations change nothing to it. — kumesana, Jul 20 '18 at 13:53

score 0 · Answer 1 · answered Jul 20 '18 at 13:42

0

The ++ operator is evaluated before the expression.

i.e.:

int x = 10;
int y = ++x; //y = 11
int z = x ++; // z = 11;

answered Jul 20 '18 at 13:42

Hongyu Wang

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1
10

Comments have point out the reason,so I think you had better avoid to answer question like this,due to someone might downvote it. – flyingfox Jul 20 '18 at 13:43
The `++` operator is evaluated first, while `++` is evaluated *later*. – Luiggi Mendoza Jul 20 '18 at 13:44

score 0 · Answer 2 · answered Jul 20 '18 at 13:44

y-- is higher up on the list from your source. However, when the post-decrement happens, it happens after the whole evaluation.

So if you print y after getting the value of z, it will be 9.

And the pre-increment happens first, so ++x becomes 6 within that statement (and obviously multiplies by 10).

See an example in the docs.

Priority operators in Java

2 Answers2