int x=1;
int y=2;
x ^= y ^= x ^= y;
I am expecting the values to be swapped.But it gives x=0 and y=1. when i tried in C language it gives the correct result.
Asked
Active
Viewed 3,630 times
14
Hungry Blue Dev
- 1,313
- 16
- 30
sadananda salam
- 845
- 1
- 7
- 12
-
4It's undefined behavior in C, since you're modifying both x and y twice in a sequence point. – Matthew Flaschen Oct 02 '10 at 08:32
-
3Dont use it. Using an extra temp variable to swap 2 is MUCH more efficient because it doesnt have to make the computation. – Imre L Oct 02 '10 at 08:35
-
1@Imre: I will take your suggestion. – sadananda salam Oct 02 '10 at 08:39
-
2Behaviour of the expression `x ^= y ^= x ^= y;` is [undefined in C](http://stackoverflow.com/questions/3741440/potential-problem-in-swapping-values-of-two-variables-without-using-a-third-vari/3741450#3741450). – Prasoon Saurav Oct 02 '10 at 08:40
-
Sorry, that should read twice between sequence points. – Matthew Flaschen Oct 02 '10 at 08:53
-
@Matthew : Sorry I didn't see your comment before posting mine. – Prasoon Saurav Oct 02 '10 at 08:56
-
2IMre L: you are completely wrong. Just because the "computation" takes longer for us humans, there is no reason to assume that it will take longer than an assingment on a computer. In fact, I bet both will take exactly as long. XOR is about the most trivial operation a CPU can do. – Erich Kitzmueller Oct 02 '10 at 09:32
-
2The wording "correct" is incorrect. – Thorbjørn Ravn Andersen Oct 02 '10 at 10:23
-
@ammoQ: Have you done benchmarking? – Imre L Oct 02 '10 at 15:50
-
Imre: Yes, in C, gcc with -O3 optimization, identically fast. – Erich Kitzmueller Oct 02 '10 at 19:41
-
1In Java, assignments are faster. – Erich Kitzmueller Oct 02 '10 at 19:46
-
1Correction: in C,the ^= version appears to be a tiny bit faster on my computer, but it's on the edge of measurability. – Erich Kitzmueller Oct 02 '10 at 19:51
-
2The correct C version of what you are trying to do looks as folows `x ^= y; y ^= x; x ^= y;`. What you have in your original post produces undefined behavior. If it "works", it's only by accident. In reality, it *doesn't* work. – AnT stands with Russia Oct 02 '10 at 22:35
2 Answers
57
Your statement is roughly equivalent to this expanded form:
x = x ^ (y = y ^ (x = x ^ y));
Unlike in C, in Java the left operand of a binary operator is guaranteed to be evaluated before the right operand. Evaluation occurs as follows:
x = x ^ (y = y ^ (x = x ^ y))
x = 1 ^ (y = 2 ^ (x = 1 ^ 2))
x = 1 ^ (y = 2 ^ (x = 3))
x = 1 ^ (y = 2 ^ 3) // x is set to 3
x = 1 ^ (y = 1)
x = 1 ^ 1 // y is set to 1
x = 0 // x is set to 0
You could reverse the order of the arguments to each xor expression so that the assignment is done before the variable is evaluated again:
x = (y = (x = x ^ y) ^ y) ^ x
x = (y = (x = 1 ^ 2) ^ y) ^ x
x = (y = (x = 3) ^ y) ^ x
x = (y = 3 ^ y) ^ x // x is set to 3
x = (y = 3 ^ 2) ^ x
x = (y = 1) ^ x
x = 1 ^ x // y is set to 1
x = 1 ^ 3
x = 2 // x is set to 2
This is a more compact version that also works:
x = (y ^= x ^= y) ^ x;
But this is a truly horrible way to swap two variables. It's a much better idea to use a temporary variable.
Mark Byers
- 811,555
- 193
- 1,581
- 1,452
-
Are you sure? I thought an assignment returned true, which is why if statements where = was used instead of == always ran? That'd mean your first statement was the same as x=x^true^true; – AaronM Oct 02 '10 at 22:38
-
1@Aaron, no. JLS [§15.26](http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.26): "At run time, the result of the assignment expression is the value of the variable after the assignment has occurred." In Java, the old `=` instead of `==` in an `if` condition only compiles if the variable is a boolean. – Matthew Flaschen Oct 02 '10 at 22:54
15
Mark is completely correct about how it evaluates in Java. The reason is JLS §15.7.2., Evaluate Operands before Operation, and §15.7, which requires evaluation left to right:
It is equivalent (by §15.26.2, Compound Assignment Operators) to:
x = x ^ (y = y ^ (x = (x ^ y)));
We evaluate left to right, doing both operands before the operation.
x = 1 ^ (y = y ^ (x = (x ^ y))); // left of outer
x = 1 ^ (y = 2 ^ (x = (x ^ y))); // left of middle
x = 1 ^ (y = 2 ^ (x = (1 ^ y))); // left of inner
x = 1 ^ (y = 2 ^ (x = (1 ^ 2))); // right of inner
x = 1 ^ (y = 2 ^ (x = 3)); // inner xor (right inner assign)
x = 1 ^ (y = 2 ^ 3); // inner assign (right middle xor)
x = 1 ^ (y = 1); // middle xor (right middle assign)
x = 1 ^ 1; // middle assign (right outer xor)
x = 0; // outer xor (right outer assign)
Note that it is undefined behavior in C, because you're modifying the same variable twice between sequence points.
Matthew Flaschen
- 278,309
- 50
- 514
- 539