Java order of precedence

Question

I just came across a piece of code I find interesting (because I have never seen it as a question before in 2 years of programming)

int x = 5;
int y = 3;
int z = y + (y+=1) % 4 + (x-=2) / 3;
System.out.println(z);

The output is 4.

I am wondering why is the left most 'y' evaluated first instead of the '(y+=1)' which would then resulted in an output of 5. (in other words, why is the bracket not forcing the order of precedence?)

I am not sure what to search since searching 'java order of precedence' returns results that at best shows tricky examples of y++, ++y kind of questions or just the order of precedence table.

I tagged Java but I have tested this with C# and javascript so it is probably a general thing in programming.

Update

I was mistaken about the order of precedence and order of evaluation.

This article helped me to further understand the answers provided.

You wouldn't see that as a question because any programmer worth his salt wouldn't write something like that. These kinds of questions are quite worthless. — Kayaman
– Kayaman, Commented Aug 25, 2016 at 11:36
@JFMeier I would like to know why the bracket '(y+=1)' was not done first. — Stein121
– Stein121, Commented Aug 25, 2016 at 11:37
@Kayaman I know that no 1 would write something like this, it is just 1 of those academic questions used to test students on their understanding of order of precedence. — Stein121
– Stein121, Commented Aug 25, 2016 at 11:37
Don't confuse precedence with order of evaluation. The brackets change how the expression is parsed but not the order of evaluation. — Peter Lawrey
– Peter Lawrey, Commented Aug 25, 2016 at 11:42

Bathsheba · Accepted Answer · 2016-08-25 11:49:54Z

11

In short, the parentheses only apply to a particular term.

The expression y + (y+=1) % 4 + (x-=2) / 3; can be rewritten as t1 + t2 + t3, with t1 standing for y, t2 for (y+=1) % 4, and t3 for (x-=2) / 3.

Java always evaluates this from left to right, since the associativity of the binary operator + is from left to right. Hence t1 is the first term to be evaluated and so is done so with the unincremented value of y.

edited Aug 25, 2016 at 11:49

answered Aug 25, 2016 at 11:36

Bathsheba

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10 Comments

Boris the Spider Over a year ago

This is the key - evaluation is always LTR. I believe in C++ the evaluation order is undefined...

Bathsheba Over a year ago

Indeed, the evaluation of + is from LTR: I've added that. Note that ternary conditionals, for example, are evaluated from RTL.

Peter Lawrey Over a year ago

This is the order most western language are read left to right. Arabic, Hebrew are read right to left by comparison.

Bathsheba Over a year ago

@BoristheSpider: In C++ and C, the value of this this expression is indeed undefined. But it's more to do with the fact that + is not a sequencing point in C and C++, as opposed to the order of evaluation.

Klitos Kyriacou Over a year ago

A source of confusion is school mathematics teachers who say you should evaluate brackets first (with mnemonics such as BODMAS). Unfortunately, people tend to confuse precedence order with evaluation order, They are totally different things.

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ravthiru · Accepted Answer · 2017-02-01 05:41:46Z

8

As per Java language Specs Evaluate Left-Hand Operand First and Evaluate Operands before Operation

edited Feb 1, 2017 at 5:41

answered Aug 25, 2016 at 11:38

ravthiru

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1 Comment

Dawood ibn Kareem Over a year ago

This is the correct answer - the fact that evaluation order is prescribed by the Java Language Specification.

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