Java primitive variable reassignment during expression

Question

I had a question in my test that I got confused about (code attached below). To put it shortly, I thought that the variables are reassigned and then added back as a value to the expression (making the output "8, 10") but seems like the original value somehow is not changed. What am I missing?

p.s. Sorry if a similar question exists, I couldn't find one (probably its too obvious :P).

class InitTest{
    public static void main(String[] args){
        int a = 10;
        int b = 20;
        a += (a = 4);
        b = b + (b = 5);
        System.out.println(a + ",  " + b);
    }
}

Please do not write code like this - as you have probably noticed, it is confusing to follow what it actually is doing. — , Sep 02 '20 at 00:10
@Taschi: bad code like this is probably the cornerstone of the programming certification exams — Hovercraft Full Of Eels, Sep 02 '20 at 00:14
@HovercraftFullOfEels, I am not a certified Java dev, but if certification exams actually teach people to rely on edge cases, then it is even more necessary to point out that this is not good code. — , Sep 02 '20 at 00:17
@Taschi-corporationsarebad It's not my code. It was an example and I had to choose an answer. — Artur Shalimov, Sep 02 '20 at 00:25

Jacob G. · Accepted Answer · 2020-09-02T00:35:56.663

1

a += (a = 4);

The above is logically equivalent to the following:

a = a + (a = 4);

If we substitute in the existing value for a, then this simplifies to:

a = 10 + 4 = 14

We can do the same for b:

b = b + (b = 5) = 20 + 5 = 25

We see this result because of operator precedence. The addition operator, +, has a higher precedence than the assignment operator, =, as defined by the Java operator precedence table.

You can see that the addition-assignment operator, +=, shares the same precedence with the assignment operator, in which case the expression is evaluated from left to right.

If, instead, the expressions were:

a = (a = 4) + a;
b = (b = 5) + b;

Then it would result in the output that you expect (a = 8, b = 10), as the left operand is computed before the right operand (when evaluating the expression). I'll try to locate where this is specified in the Java Language Specification.

edited Sep 02 '20 at 00:35

answered Sep 02 '20 at 00:07

Jacob G.

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Ooooh, I got it now. My mistake was that I assumed that the value would dynamically change – Artur Shalimov Sep 02 '20 at 00:17
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Expressions are evaluated according to operator precedence. It is *operands* that are evaluated left to right. The OP's results are due to operator precedence. – user207421 Sep 02 '20 at 00:24
@JacobG.So in this case parentheses are not counted as arithmetic but just as a separator for the assignment? (i thought that parentheses have higher OP) – Artur Shalimov Sep 02 '20 at 00:43
@ArturShalimov For more information about the role of parentheses in expression evaluation, I'd look here: [If parenthesis has a higher precedence then why is increment operator solved first?](https://stackoverflow.com/q/28219423/7294647) – Jacob G. Sep 02 '20 at 00:46

score -1 · Answer 2 · answered Sep 02 '20 at 00:14

-1

In Java when assigning a value to a variable, the first thing that is calculated is the left part of the = sign and then the right. Therefore, when you write a += (a = 4); which is equivalant to a = a + (a=4) which is the same as a = a + 4 , same for b.

answered Sep 02 '20 at 00:14

Kostas Konnidis

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Why would the left part of the = sign ever be "calculated"? It's just an identifier. You can't assign something to `a + 5`. – Sep 02 '20 at 00:15
@Taschi-corporationsarebad The left side has to be resolved to a variable, which might be an object's field or an array component; for example, `getObj().getArray()[4] = 5` obviously has to call two methods. The left-hand side is evaluated before the right-hand side is evaluated. – kaya3 Sep 02 '20 at 00:40
@kaya3, fair enough, but evaluating is not the same as calculating. My example of `a + 5` is not a valid left side for an assignment. – Sep 02 '20 at 00:46
Whether a computation is or is not a "calculation" is not really defined in a technical sense. But if it has to do arithmetic with numbers to count as a calculation, then `foo(a + 5).bar = "baz";` should meet your criterion. – kaya3 Sep 02 '20 at 11:28

Java primitive variable reassignment during expression

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