无法理解 Java 中的按位 & 运算符

Question

int number = 3;
System.out.println(number & 1 << 2);

Given this snippet where I am performing bitwise AND to number and then left shifting by 2, Why is the result 0 and not 4 (0100)?

Answer 1

让它(number & 1) << 2因为你正在做的意味着移动 1 << 2 次，然后 & 与数字。

Answer 2

First, as multiple people have already mentioned, the & operator's precedence is smaller than <<'s. So your code would look like this: number & (1 << 2) . Let's imagine your code looked like this instead: (number % 1) << 2 . Now, what does this do?

First, let's talk about the & operator. What does it do? In short, it will apply an AND gate to each bit of the two given numbers, recording the result of the gate in a new number:

a      0 0 1 1
b      1 1 1 0
result 0 0 1 0

An AND gate works in the following way: the result of this gate is 1 only when both inputs are 1, otherwise, the output of the gate is 0.

In your case, you have the following:

a         ...number
b   0 0 0 0 0 0 0 1

Since each bit but the first one of b is 0, the result will be all 0s, except for the first bit, which will be whatever the first bit of number was (note that a & 1 is virtually equivalent to a % 2 ).

The shift operator now will shift the only remaining bit to the left 2 bits, which will virtually multiply it by 4.

So, for example, if number was 3, 3 & 1 would be 1 (the first bit of 3 is 1), and then 1 would be shifted over 2 bits, so the result would be 4.

In fact, the expression (number & 1) << 2 will produce only two values:

• 4, when the number is odd (its first bit is 1)
• 0, when the number is even (its first bit is 0)