对于AND运算符和OR运算符，Java有两类求值，即Short-Circuit求值和full求值。

&& ||短路评估

短路求值使您可以不计算AND和OR表达式的右边，当整体结果可以从左边的值预测时。

int numberOne = 1;
int numberTwo = 2;
boolean result = false;

// left-side is false so the the overall result CAN be predicted without evaluating the right side.
// numberOne will be 1, numberTwo will be 2, result will be false
result = (numberOne > numberTwo) && (++numberOne == numberTwo);

System.out.println(numberOne); // prints 1
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints false


// left-side is true so the the overall result CAN NOT be predicted without evaluating the right side.
// numberOne will be 2, numberTwo will be 2, result will be true
result = (numberTwo > numberOne) && (++numberOne == numberTwo);

System.out.println(numberOne); // prints 2
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints true

^全面评估

虽然在某些情况下可以预测结果，但有必要计算右边的值。

int numberOne = 1;
int numberTwo = 2;
boolean result = false;

// left-side is false so the the overall result will be false BUT the right side MUST be evaluated too.
// numberOne will be 2, numberTwo will be 2, result will be false
result = (numberOne > numberTwo) & (++numberOne == numberTwo);

System.out.println(numberOne); // prints 2
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints false

注意:

Notice that for XOR (^) there is no short-circuit, because both sides are always required to determine the overall result. Notice that other possible names for Short-Circuit evaluation are minimal evaluation and McCarthy evaluation. It is not recommenced to mix boolean logic and actions in the same expression & can also act as a Bitwise AND operator which is very academic and can be used in cryptography. When both bits are 1, the result is 1, or either of the bits is not 1, the result is 0. (Check the following code)

AND位的例子:

byte a = 5;              // 00000101
byte b = 3;              // 00000011
byte c = (byte) (a & b); // 00000001 (c is 1)

除了&&和||是短路外，在混合这两种形式时，还要考虑运算符优先级。 我认为每个人都不会立即看出result1和result2包含不同的值。

boolean a = true;
boolean b = false;
boolean c = false;

boolean result1 = a || b && c; //is true;  evaluated as a || (b && c)
boolean result2 = a  | b && c; //is false; evaluated as (a | b) && c

如JLS(15.22.2)中所述:

When both operands of a &, ^, or | operator are of type boolean or Boolean, then the type of the bitwise operator expression is boolean. In all cases, the operands are subject to unboxing conversion (§5.1.8) as necessary. For &, the result value is true if both operand values are true; otherwise, the result is false. For ^, the result value is true if the operand values are different; otherwise, the result is false. For |, the result value is false if both operand values are false; otherwise, the result is true.

“诀窍”在于&是一个整数位运算符，也是一个布尔逻辑运算符。为什么不呢，把这个作为运算符重载的例子是合理的。

boolean a, b;

Operation     Meaning                       Note
---------     -------                       ----
   a && b     logical AND                    short-circuiting
   a || b     logical OR                     short-circuiting
   a &  b     boolean logical AND            not short-circuiting
   a |  b     boolean logical OR             not short-circuiting
   a ^  b     boolean logical exclusive OR
  !a          logical NOT

short-circuiting        (x != 0) && (1/x > 1)   SAFE
not short-circuiting    (x != 0) &  (1/x > 1)   NOT SAFE

对于AND运算符和OR运算符，Java有两类求值，即Short-Circuit求值和full求值。

&& ||短路评估

短路求值使您可以不计算AND和OR表达式的右边，当整体结果可以从左边的值预测时。

int numberOne = 1;
int numberTwo = 2;
boolean result = false;

// left-side is false so the the overall result CAN be predicted without evaluating the right side.
// numberOne will be 1, numberTwo will be 2, result will be false
result = (numberOne > numberTwo) && (++numberOne == numberTwo);

System.out.println(numberOne); // prints 1
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints false


// left-side is true so the the overall result CAN NOT be predicted without evaluating the right side.
// numberOne will be 2, numberTwo will be 2, result will be true
result = (numberTwo > numberOne) && (++numberOne == numberTwo);

System.out.println(numberOne); // prints 2
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints true

^全面评估

虽然在某些情况下可以预测结果，但有必要计算右边的值。

int numberOne = 1;
int numberTwo = 2;
boolean result = false;

// left-side is false so the the overall result will be false BUT the right side MUST be evaluated too.
// numberOne will be 2, numberTwo will be 2, result will be false
result = (numberOne > numberTwo) & (++numberOne == numberTwo);

System.out.println(numberOne); // prints 2
System.out.println(numberTwo); // prints 2
System.out.println(result);    // prints false

注意:

Notice that for XOR (^) there is no short-circuit, because both sides are always required to determine the overall result. Notice that other possible names for Short-Circuit evaluation are minimal evaluation and McCarthy evaluation. It is not recommenced to mix boolean logic and actions in the same expression & can also act as a Bitwise AND operator which is very academic and can be used in cryptography. When both bits are 1, the result is 1, or either of the bits is not 1, the result is 0. (Check the following code)

AND位的例子:

byte a = 5;              // 00000101
byte b = 3;              // 00000011
byte c = (byte) (a & b); // 00000001 (c is 1)

& <——验证两个操作数 && <——如果第一个操作数求值为false，则停止求值，因为结果将为false

(x != 0) & (1/x > 1) <——这意味着求(x != 0)然后求(1/x > 1)然后执行&。问题是当x=0时，这会抛出异常。

(x != 0) && (1/x > 1) <——这意味着求(x != 0)，只有当它为真时，才求(1/x > 1)，所以如果你有x=0，那么这是完全安全的，并且不会抛出任何异常，如果(x != 0)求值为假，整个事情直接求值为假，而不求(1/x > 1)。

编辑:

exprA | exprB <——这意味着评估exprA然后评估exprB，然后执行|。

exprA || exprB <——这意味着评估exprA，只有当这是假的，然后评估exprB，并执行||。