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.
2 Answers
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.
2 Comments
AaronM
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;
Matthew Flaschen
@Aaron, no. JLS §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.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.
x ^= y ^= x ^= y;is undefined in C.