a^=b^=a^=b in Java seems wrong

Question

public class Main
{
    public static void main(String[] args)
    {
        int a,b,c,d;
        a=1;b=2;
        a^=b^=a^=b;
        System.out.println(a+" "+b);
        c=1;d=2;
        c^=d;
        d^=c;
        c^=d;
        System.out.println(c+" "+d);
    }
}

I use a^b^=a^=b to swap a and b. However the output of this program is

0 1
2 1

I am a new one in Java, and I don't know why a is 0. Is it a bug of my java runtime environment? Or there are something special in Java I don't know? Here is java -version.

java version "1.7.0_65"
OpenJDK Runtime Environment (fedora-2.5.2.5.fc20-x86_64 u65-b17)
OpenJDK 64-Bit Server VM (build 24.65-b04, mixed mode)

Sorry it's my first question on stack overflow ... if I forgot some rules, tell me please, thank you.

My advice? Don't write code like this. It's an obfuscation code winner. Your goal should be to write as clearly as possible. — duffymo, Oct 09 '14 at 12:30
I know it's not a good way to code, but I wonder what happened when I used a^=b^=a^=b. I don't think a good programmer will do it too. — Sweetdumpling Lin, Oct 09 '14 at 12:31
It's perfectly "legal" code, from a compiler standpoint, but I've got 40 years of programming experience, 10 with Java, and I couldn't tell you what it means. **Don't do it.** — Hot Licks, Oct 09 '14 at 12:32
If you really want to know what it means, refer to the [JLS](http://docs.oracle.com/javase/specs/jls/se7/html/index.html) and work it out. It would be a good "exercise" in deciphering cryptic code. — Hot Licks, Oct 09 '14 at 12:34
Well, I know what it means. What I wonder what's the difference between what I do on a/b and c/d. It seems same but the answer is different. — Sweetdumpling Lin, Oct 09 '14 at 12:36
I think the purpose of the example is to demonstrate that such complex expression can bite you in the butt. — Hot Licks, Oct 09 '14 at 12:38
I gave this question a down-vote, since I feel this strays beyond the type of theoretical questions that are welcome here into the *not useful* category. I think we all agree such code won't be written "for real", so having various people chew over it for a while to see what it means doesn't seem like a great idea. — Duncan Jones, Oct 09 '14 at 12:39
(BTW, though it's a cute parlor trick it's foolish/inefficient (and in a few odd cases dangerous) to use the old 3 XOR scheme to swap values, except in the very specific situation of assembly language when the number of registers is limited. Use a temp.) — Hot Licks, Oct 09 '14 at 12:42
If you really want to do it in one line without temp variables, use a+=(b-(b=a)); — Nuri Tasdemir, Oct 09 '14 at 12:50

score 4 · Accepted Answer · edited May 23 '17 at 12:11

4

Let's simplify it together

int a = 1;
int b = 2;
a ^= (b ^= (a ^= b));
System.out.println(a + " " + b);

is the same as saying

int a = 1;
int b = 2;
int a_tmp = a;
a = a_tmp ^ (b ^= (a ^= b));
System.out.println(a + " " + b);

Your code is assuming that we process a ^= (b ^= (a ^= b)) before a ^= (b ^= (a ^= b)) but that's wrong. I don't really know where it will work. But as pointed by @duffymo don't try this at home.

edited May 23 '17 at 12:11

Community

1
1

answered Oct 09 '14 at 12:42

talex

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Yep, *any* statement with multiple assignments is dangerous at best. And the "operator-equals" operators are doubly dangerous. – Hot Licks Oct 09 '14 at 12:44

score 0 · Answer 2 · answered Oct 09 '14 at 13:35

Make sure you look at your order of operations. This may be what is happening:

true ^ true = false  
true ^ false = true 
false ^ true = true 
false ^ false = false

a^=b^=a^=b;

Work right to left: 0001 ^ 0010 ^ 0001 ^ 0010 take the first 2 farthest right 0001 ^ 0010 = 0011. Now keep moving left take your answer with the next element 0010 ^ 0011 = 0001 now move left one more time. 0001 ^ 0001 = 0000 so a = 0000 or 0.
Now take the b expression: 0010 ^ 0001 ^ 0010 now do the same thing right to left. 0001 ^ 0010 = 0011 now move to the left 0010 ^ 0011 = 0001 so b = 0001 or 1. So when you print a = 0000 or 0 and b = 0001 or 1. This is bitwise exclusive OR operations make sure to look at order of operations carefully.

Your explanation is wrong. You assume that the values of the variables haven't changed yet by putting in 0001 ^ 0010 ^ 0001 ^ 0010 as the expression to evaluate. But actually it is a ^ 0010 ^ 0001 ^ 0010. Because after the evaluation of the first xor 0001 ^ 0010 also changes a. So it no longer is 0001 but 0011 for the first variable in your statement. The OP is actually correct by expecting 2 and 1 as values (according to the specs). I will post an answer later with why the result is 0 and 1 instead. — Juru, Oct 10 '14 at 06:43

a^=b^=a^=b in Java seems wrong

2 Answers2