为什么这个语句不适用于java x ^ = y ^ = x ^ = y;

Question

int x=1;
int y=2;
x ^= y ^= x ^= y;

我期待值被交换.但它给出x = 0和y = 1.当我尝试使用C语言时,它会给出正确的结果.

Answer 1

您的陈述大致相当于此扩展形式:

x = x ^ (y = y ^ (x = x ^ y));

与C不同,在Java中,保证在右操作数之前计算二元运算符的左操作数.评估如下:

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

您可以反转每个xor表达式的参数顺序,以便在再次计算变量之前完成赋值:

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

这是一个更紧凑的版本,也可以工作:

x = (y ^= x ^= y) ^ x;

但这是交换两个变量的真正可怕方式.使用临时变量是一个更好的主意.

Answer 2

Mark对于如何在Java中进行评估是完全正确的.原因是JLS§15.7.2.,在操作前评估操作数,以及 §15.7,这需要从左到右进行评估:

它等同于(通过§15.26.2,复合赋值运算符):

x = x ^ (y = y ^ (x = (x ^ y)));

我们从左到右评估,在操作之前执行两个操作数.

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)

请注意,它在C中是未定义的行为,因为您在序列点之间修改了相同的变量两次.