Assigning a variable during arithmetic in Java?

Question

My professor gave us this java snippet during a lecture and I don't understand why it outputs 12.

int b = 9; 
b = b + (b = 3); 
System.out.println( "b = " + b );

My thinking is that since parentheses give operations precedence, b would be reassigned to 3 first. Then, it would go b = 3 + 3 = 6. Obviously, this isn't the case and b = 9 + 3 = 12 is actually executed. Why is the code executed left to right and not parentheses first?

Excuse my poor title, I'm not sure what to call this because I don't think you ever actually program this way.

Answer 1

The entire expression on the right hand side is evaluated before the value of b is then changed. Also, it is evaluated from left to right. So the value of b does change during the evaluation of the expression, but the first b has already been sampled as 9 before the second term sets b to 3. It doesn't matter that b is eventually being set to the result of the expression. The term (b = 3) has the value 3. So the assignment is just:

b = 9 + 3

And so the result you get is 12.

On the other hand, the statement:

b = (b = 3) + b;

simplifies to

b = 3 + 3

Resulting in 6. This occurs because, again, the expression is evaluated left to right. So the first term has a value of 3 and also sets b to 3 before the second term b is then evaluated, and so has a value of 3.

Answer 2

Precedence does not mean that it runs first. It merely lets you rewire operator precedence. Ordinarily something like 1 + 2 * 3 is resolved as:

• A plus operation between A and B, where
• A is the integer literal '1'
• B is the expression 2 * 3.

Why? Because operator precedence rules state that * binds tighter than +. With parens you can override that; (1 + 2) * 3 is a multiply operation between A and B where B is a '3' literal and A is '1 + 2'. That's all it does. It doesn't change the order in which things are resolved.

Java first evaluates 'b' (it's 9), then it evaluates (b = 3) (which is 3, and as a side effect, makes b become 3), and thus that expression is 12, which is then assigned to b, so it's 12.

Why does it evaluate the left hand side of that + first? Because the spec says so: resolves left to right.

Try this for funsies:

int b = 9;
b = (b = 3) + b;
System.out.println(b);

now it prints 6 as expected.