将长度为3的字节数组转换为int

Question

I have a byte array of length 3, representing a decimal binary number. My question:

Why is this correct:

(a) int x = (array[0] & 0xff) << 16 | (array[1] & 0xff) << 8 | (array[2] & 0xff); int x = (array[0] & 0xff) << 16 | (array[1] & 0xff) << 8 | (array[2] & 0xff);

but this isn't?

(b) int x = array[0] << 16 | array[1] << 8 | array[2]; int x = array[0] << 16 | array[1] << 8 | array[2];

Let's say array[0] is 01010101 . Isn't this what happens?

array[0] & 0xff = 01010101 & 11111111 = 01010101 = array[0]

Why is option b) wrong?

Answer 1

Consider what happens when the most significant bit of array[0] is 1.

For example:

array[0] = (byte)0xff;
System.out.println (array[0] << 16);
System.out.println ((array[0] & 0xff) << 16);

output:

-65536
16711680

array[0] is converted to an int for the sake of the left-shift operator. If it has a negative value as a byte , it will have a negative value as an int , and will have a negative value after the left-shift.

When you perform bit-wise AND with 0xff , you make sure the result will be positive.