Java中&=和|=短路吗？
换句话说，以下两个语句的行为是否相同？
isFoobared = isFoobared && methodWithSideEffects();
isFoobared &= methodWithSideEffects();
我意识到我可以编写一个测试，但有人可能会立即知道这一点，而其他人可能会发现答案有用。
In other words, do the following two statements behave the same way?
isFoobared = isFoobared && methodWithSideEffects();
isFoobared &= methodWithSideEffects();
I realize I could just write up a test, but someone might know this offhand, and others might find the answer useful.
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评论（2）
不，
|=和&=不会短路，因为它们是&和|，不短路。因此，假设
boolean &，isFoobared &= methodWithSideEffects()的等价物是：另一方面
&&和 < code>|| 做了短路，但令人费解的是，Java 没有它们的复合赋值版本。也就是说，Java 既没有&&=也没有||=。另请参阅
这个短路业务到底是什么？
boolean逻辑运算符（&和|）与其对应的boolean条件运算符（>&&和||) 是前者不会“短路”；后者确实如此。也就是说，假设没有例外等：&和|总是 计算两个操作数&&和 < code>|| 有条件地计算右操作数；仅当右操作数的值可能影响二元运算的结果时，才会计算右操作数。这意味着在以下情况下不会评估正确的操作数：&&的左操作数的计算结果为falsefalse）||的左操作数计算结果为truetrue）参考文献
&、^和|< /a>& ;&||No,
|=and&=do not shortcircuit, because they are the compound assignment version of&and|, which do not shortcircuit.Thus, assuming
boolean &, the equivalence forisFoobared &= methodWithSideEffects()is:On the other hand
&&and||do shortcircuit, but inexplicably Java does not have compound assignment version for them. That is, Java has neither&&=nor||=.See also
What is this shortcircuiting business anyway?
The difference between the
booleanlogical operators (&and|) compared to theirbooleanconditional counterparts (&&and||) is that the former do not "shortcircuit"; the latter do. That is, assuming no exception etc:&and|always evaluate both operands&&and||evaluate the right operand conditionally; the right operand is evaluated only if its value could affect the result of the binary operation. That means that the right operand is NOT evaluated when:&&evaluates tofalsefalse)||evaluates totruetrue)References
&,^, and|&&||不，他们没有，因为
x &= y是x = x & 的缩写。 y和x |= y是x = x | 的缩写。 y 。 Java 没有&&=或||=运算符来执行您想要的操作。&和|运算符（以及~、^、<<、>和>>>) 是 按位运算符。表达式x & y对于任何整型类型，都会执行按位与运算。类似地，|执行按位或。要执行按位运算，数字中的每一位都被视为布尔值，1表示true，0表示false< /代码>。因此，3 & 2 == 2，因为3在二进制中是0...011，而2是0... 010。类似地，3 | 2 == 3 。 Wikipedia 对不同的运算符有很好的完整解释。现在，对于布尔值，我认为您可以使用&和|作为非短路 em> 相当于&&和||，但我无法想象你为什么想要这样做。No, they do not, because
x &= yis short forx = x & yandx |= yis short forx = x | y. Java has no&&=or||=operators which would do what you want.The
&and|operators (along with~,^,<<,>>, and>>>) are the bitwise operators. The expressionx & ywill, for any integral type, perform a bitwise and operation. Similarly,|performs a bitwise or. To perform a bitwise operation, each bit in the number is treated like a boolean, with1indicatingtrueand0indicatingfalse. Thus,3 & 2 == 2, since3is0...011in binary and2is0...010. Similarly,3 | 2 == 3. Wikipedia has a good complete explanation of the different operators. Now, for a boolean, I think you can get away with using&and|as non-short-circuiting equivalents of&&and||, but I can't imagine why you'd want to anyway.