Why the need of a bitwise "and" for some char to byte conversions in Java?

Question

Here I'm working with a Java to C# sample app translation that involves cryptography (AES and RSA and so on...)

At some point in Java code (the one that actually works and being translated to C#), I've found this piece of 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 googling (here), I've seen that this is a common behaviour mainly on Java code...

I know that char is a 16-bit type and byte is 8-bit only, but I couldn't understand the reason for this bitwise and operation for a char->byte conversion.

Could someone explain?

Thank you in advance.

Answer 1

In this case, it is quite unnecessary, but the sort of thing people put in "just in case". According to the JLS, 5.1.3. Narrowing Primitive Conversion

A narrowing conversion of a char to an integral type T likewise simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the resulting value to be a negative number, even though chars represent 16-bit unsigned integer values.

Similar code is often needed in widening conversions to suppress sign extension.

Answer 2

When you convert 0x00FF to binary it becomes 0000 0000 1111 1111

When you and anything with 1, it is itself:

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

When you and anything with 0, it is 0

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

When this operation occurrs encodedArr[j] = (byte) (data[i] & 0x00FF); it's taking the last 8 bits and the last 8 bits only of the data and storing that. It is throwing away the first 8 bits and storing the last 8

The reason why this is needed is because a byte is defined as an eight bit value. The bitwise and exists to stop a potential overflow -> IE assigning 9 bits into a byte

A char in Java is 2 bytes! This logic is there to stop an overflow. However, as someone pointed out below, this is pointless because the cast does it for you. Perhaps someone was being cautious?

Answer 3

It's a way to truncate the value by keeping only the least significat bits so it "fits" in a byte!

I hope it helps!