Doubt in Compound Operator Evaluation

[Sidharth Khattri]

I can't figure out how the compound operator are evaluated. Like in this example, I thought the answer would be "s 17" but it is "s 16"
++x i.e 6 is passed to dostuff() and:
s += EASY + ++s;
would be:
EASY + ++s = 3 + 7 = 10
So, s += 10
as s is already incremented to 7, shouldn't s+= 10 be s = 7+10? i.e s = 17? Why doesn't it happen this way? What is the actual rule for evaluation?

public class Test {
public static final int EASY = 3;
public static void main(String [] args) {
int x = 5;
new Test().doStuff(++x);
}
public void doStuff(int s) {
s += EASY + ++s;
System.out.println("s " + s);
}
}

Output is: s 16

Similarly, how is this expression evaluated?

int a = 3, c = 1;
a += (a += c)	//Output is: 7

I guessed 8.
Can anyone please explain the rules followed in these compound operations?

[Henry Wong]

Sidharth Khattri wrote:
Similarly, how is this expression evaluated?

int a = 3, c = 1;
a += (a += c)	//Output is: 7

I guessed 8.
Can anyone please explain the rules followed in these compound operations?

Expression evaluations are done, as stated in the Java Language Specification, as follows...

http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.7

Basically, to summarize, they are evaluated left-to-right, with side-effects taking place as soon as the operator is used (completed).

So, in this example, the first compound operator evaluates "a" to be the sum of 3 plus the result of "a" after the second compound operator completes. Granted "a" is no longer three (by the time the second compound operator completes), but that part of the expression has already been evaluated.

Henry

[Chan Ag]

s += EASY + ++s;

evaluates to s = s + EASY + ++s
= 6 + EASY + ++s
= 6 + 3 + ++s
= 9 + ++s
= 9 + 7
= 16

a += (a += c)

evaluates to
a = a + ( a+= c)
= a + ( a = a + c) // the bracket here makes all the difference.
= a + ( a = 3 + 1 )
= a + ( a =4 )
= 4 + 4
= 8

Edit : This is wrong cause s += (EASY + ++s ) ; also gives me 16. .

[Sidharth Khattri]

Henry Wong wrote:

Sidharth Khattri wrote:
Similarly, how is this expression evaluated?

int a = 3, c = 1;
a += (a += c)	//Output is: 7

I guessed 8.
Can anyone please explain the rules followed in these compound operations?

Expression evaluations are done, as stated in the Java Language Specification, as follows...

http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.7

Basically, to summarize, they are evaluated left-to-right, with side-effects taking place as soon as the operator is used (completed).

So, in this example, the first compound operator evaluates "a" to be the sum of 3 plus the result of "a" after the second compound operator completes. Granted "a" is no longer three (by the time the second compound operator completes), but that part of the expression has already been evaluated.

Henry

Thanks henry. One more question if I may, is everything in java evaluated from left-to-right? I've a memory of something about right-to-left somewhere, hence I got confused.

[Sidharth Khattri]

Chan Ag wrote:s += EASY + ++s;

evaluates to s = s + EASY + ++s
= 6 + EASY + ++s
= 6 + 3 + ++s
= 9 + ++s
= 9 + 7
= 16

a += (a += c)

evaluates to
a = a + ( a+= c)
= a + ( a = a + c) // the bracket here makes all the difference.
= a + ( a = 3 + 1 )
= a + ( a =4 )
= 4 + 4
= 8

Hi Chang,
int a = 3, c = 1;
a += (a += c) //Output is: 7 not 8.

a = a + ( a+= c)
= 3 + ( a = 3 + 1) // the bracket here makes all the difference.
= 3 + ( a =4 )
= 3 + 4
= 7

Though the brackets are evaluated first, so "a" should be 4, but as the expression is already expanded as = 3 + ( a = 3 + 1) according to what Henry said, that correct answer would be 7. The brackets part can be confusing though.

[Chan Ag]

Yes, I just realized that. Thanks for the correction.

I have edited the response with the correction that it's a wrong response. Please ignore it.

[Henry Wong]

Sidharth Khattri wrote:
Thanks henry. One more question if I may, is everything in java evaluated from left-to-right? I've a memory of something about right-to-left somewhere, hence I got confused.

You are probably referring to associativity. Some operators, including the compound assignment operators, do have right-to-left associativity. Remember that precedence, associativity, and order of evaluation, are three different things. And you will need to take all three into account. The example in this topic used parenthesis to not need to worry about precedence and associativity -- and allowed it to focus on evaluation order alone.

Henry

[Sidharth Khattri]

Sidharth Khattri wrote:

Henry Wong wrote:

Sidharth Khattri wrote:
Similarly, how is this expression evaluated?

int a = 3, c = 1;
a += (a += c)	//Output is: 7

I guessed 8.
Can anyone please explain the rules followed in these compound operations?

Expression evaluations are done, as stated in the Java Language Specification, as follows...

http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.7

Basically, to summarize, they are evaluated left-to-right, with side-effects taking place as soon as the operator is used (completed).

So, in this example, the first compound operator evaluates "a" to be the sum of 3 plus the result of "a" after the second compound operator completes. Granted "a" is no longer three (by the time the second compound operator completes), but that part of the expression has already been evaluated.

Henry

Thanks henry. One more question if I may, is everything in java evaluated from left-to-right? I've a memory of something about right-to-left somewhere, hence I got confused.

Oh, now I remember, I was confusing initialization with expression evaluation. If anyone would wonder what I was talking about:

class test{
public static void main(String[]args){
int [] i = new int[3];
i[4] = i[3]/0;
}
}

this will throw ArithmeticException and not ArrayIndexOutOfBoundsException.
And associativity - right-to-left.

[Chan Ag]

Actually I misread Henry's response earlier. So I was trying to work it until now. I re read his response and now I have the clear answer. Thanks so much.

Sorry for the confusion.