Why do you need to “and” some character conversions in Java bit by bit?

Here, I am using java to c# sample application translation, involving encryption (AES, RSA, etc...)

At some point in the Java code (actually working and converted to c#), I found this Code:

for (i = i; i < size; i++) {
    encodedArr[j] = (byte) (data[i] & 0x00FF);
    j++;
} // where data variable is a char[] and encodedArr is a byte[]

After some Google search (here), I found that this is a common behavior mainly in Java code

I know that char is a 16 bit type with only 8 bytes, but I can't understand the reason for this bit by bit and the operation of char - > byte conversion

Can anyone explain

Thank you first

Solution

When you convert 0x00ff to binary, it becomes 0000 0000 1111 1111

When you and anything with a 1, it itself is:

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

When you and any 0, it is 0

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

When this operation occurs, encodedarr [J] = (byte) (data [i]& 0x00ff); It only occupies the last 8 bits and the last 8 bits of data and stores it It discards the first 8 bits and stores the last 8 bits

This is needed because a byte is defined as an 8 - bit value Bitwise and present to prevent potential overflow – > ie allocates 9 bits to a byte

Char in Java is 2 bytes! This logic can prevent overflow However, as someone pointed out below, this is meaningless because the actor did it for you Maybe someone is cautious?