What does this boolean “(number & 1) == 0” mean?

问题

On CodeReview I posted a working piece of code and asked for tips to improve it. One I got was to use a boolean method to check if an ArrayList had an even number of indices (which was required). This was the code that was suggested:

private static boolean isEven(int number)
{
    return (number & 1) == 0;
}

As I've already pestered that particular user for a lot of help, I've decided it's time I pestered the SO community! I don't really understand how this works. The method is called and takes the size of the ArrayList as a parameter (i.e. ArrayList has ten elements, number = 10).

I know a single & runs the comparison of both number and 1, but I got lost after that.

The way I read it, it is saying return true if number == 0 and 1 == 0. I know the first isn't true and the latter obviously doesn't make sense. Could anybody help me out?

Edit: I should probably add that the code does work, in case anyone is wondering.

回答1:

Keep in mind that "&" is a bitwise operation. You are probably aware of this, but it's not totally clear to me based on the way you posed the question.

That being said, the theoretical idea is that you have some int, which can be expressed in bits by some series of 1s and 0s. For example:

...10110110

In binary, because it is base 2, whenever the bitwise version of the number ends in 0, it is even, and when it ends in 1 it is odd.

Therefore, doing a bitwise & with 1 for the above is:

...10110110 & ...00000001

Of course, this is 0, so you can say that the original input was even.

Alternatively, consider an odd number. For example, add 1 to what we had above. Then

...10110111 & ...00000001

Is equal to 1, and is therefore, not equal to zero. Voila.

回答2:

You can determine the number either is even or odd by the last bit in its binary representation:

1 -> 00000000000000000000000000000001 (odd)
2 -> 00000000000000000000000000000010 (even)
3 -> 00000000000000000000000000000011 (odd)
4 -> 00000000000000000000000000000100 (even)
5 -> 00000000000000000000000000000101 (odd)
6 -> 00000000000000000000000000000110 (even)
7 -> 00000000000000000000000000000111 (odd)
8 -> 00000000000000000000000000001000 (even)

& between two integers is bitwise AND operator:

0 & 0 = 0
0 & 1 = 0
1 & 0 = 0
1 & 1 = 1

So, if (number & 1) == 0 is true, this means number is even.

Let's assume that number == 6, then:

6 -> 00000000000000000000000000000110 (even)

     &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

1 -> 00000000000000000000000000000001

-------------------------------------

0 -> 00000000000000000000000000000000

and when number == 7:

7 -> 00000000000000000000000000000111 (odd)

     &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

1 -> 00000000000000000000000000000001

-------------------------------------

1 -> 00000000000000000000000000000001

回答3:

& is the bitwise AND operator. && is the logical AND operator

In binary, if the digits bit is set (i.e one), the number is odd.

In binary, if the digits bit is zero , the number is even.

(number & 1) is a bitwise AND test of the digits bit.

Another way to do this (and possibly less efficient but more understandable) is using the modulus operator %:

private static boolean isEven(int number)
{
    if (number < 0)
       throw new ArgumentOutOfRangeException();

    return (number % 2) == 0;
}

回答4:

This expression means "the integer represents an even number".

Here is the reason why: the binary representation of decimal 1 is 00000000001. All odd numbers end in a 1 in binary (this is easy to verify: suppose the number's binary representation does not end in 1; then it's composed of non-zero powers of two, which is always an even number). When you do a binary AND with an odd number, the result is 1; when you do a binary AND with an even number, the result is 0.

This used to be the preferred method of deciding odd/even back at the time when optimizers were poor to nonexistent, and % operators required twenty times the number of cycles taken by an & operator. These days, if you do number % 2 == 0, the compiler is likely to generate code that executes as quickly as (number & 1) == 0 does.

回答5:

Single & means bit-wise and operator not comparison

So this code checks if the first bit (least significant/most right) is set or not, which indicates if the number is odd or not; because all odd numbers will end with 1 in the least significant bit e.g. xxxxxxx1

回答6:

& is a bitwise AND operation.

For number = 8:

The result is that (8 & 1) == 0. This is the case for all even numbers, since they are multiples of 2 and the first binary digit from the right is always 0. 1 has a binary value of 1 with leading 0s, so when we AND it with an even number we're left with 0.

回答7:

The & operator in Java is the bitwise-and operator. Basically, (number & 1) performs a bitwise-and between number and 1. The result is either 0 or 1, depending on whether it's even or odd. Then the result is compared with 0 to determine if it's even.

Here's a page describing bitwise operations.