Arithmetic Operators in Java (Beginner Question)

Question

I know that array operators have the precedence. Then the binary arthimetic operators * , / , % . Then + and - which they are low precedence.

But I'm confused which one will java solve first in this example. And if we have 2 operators have the same priority, what operator will be used first in java?

Thank you.

int x = y = -2 + 5 * 7 - 7 / 2 % 5;

If someone could solve this for me and explain to me part by part. Because this always confuses me in exams.

Answer 1

If operators have the same precedence then they are evaluated from left to right.

From the tutorial :

When operators of equal precedence appear in the same expression, a rule must govern which is evaluated first. All binary operators except for the assignment operators are evaluated from left to right; assignment operators are evaluated right to left.

In the expression, 7 / 2 % 5 , the / and % have the same precedence, so going left to right 7 / 2 = 3 and 3 % 5 = 3 .

The highest precedence is given to * / %. Here is the breakdown of your example:

  -2 + 5 * 7 - 7 / 2 % 5
= -2 + (5 * 7) - (7 / 2 % 5)
= -2 + 35 - (3 % 5)
= -2 + 35 - 3
= 30

Answer 2

y will be assigned the value of -2 + 5 * 7 - 7 / 2 % 5 . Then x will be assigned y 's value.

Arithmetic expression will be evaluated like:

-2 + (5 * 7) - ((7 / 2) % 5)

Here's an explanation of Java's operator precedence.

Answer 3

This looks like what you need to read: Java Operators Tutorial .

Read the tutorial and then write yourself an example program and play around with it until you're happy with operator precedence. It's the best way to learn.

Answer 4

int x = y = -2 + 5 * 7 - 7 / 2 % 5;

is same as

int x = y = (-2 + ((5 * 7) - ((7 / 2) % 5)));

/,* and % (multiplicative) are having same precedence and their association is left to right.
+ and - (additive)are having same precedence and their association is left to right.
multiplicative operations are higher precedence over additive operations.

Answer 5

Not entirely related, but you may find this of interest. It is to do with sequence points, which are points in a program which are essentially the points where the compiler ensures that everything is synchronised. It arose on SO because of the question what does

x = x++; // Operator prcedence and/or sequence point problem;

do? Or worse

x[i]=i++ + 1;// sequence point problem

Why does this go into an infinite loop?

http://www.angelikalanger.com/Articles/VSJ/SequencePoints/SequencePoints.html