Explicit conversion from int to byte in Java

Question

I am trying to convert an int to byte.

int i = 128;
byte b = (byte) i;

I know the range of byte if -128 to 127 and the rule of storing an i

Answer 1

You can simply use the following method:

byteValue()

This returns the value of this Integer as a byte.

You will need to either import Java.lang or type it out as:

Java.lang.Integer.byteValue() 

Answer 2

Bytes are treated as signed values in Java; like you pointed out, the range of a byte is [-128,127]. Adding 1 to 127 flips the sign bit, which means that 127 + 1 = -128.

Answer 3

Casting to byte doesn't mean int_value % byte_range. It means that only the last 8 significant bits are kept in the byte value. At first that seems to mean the same thing, but bytes are signed.

The int value 128 is the following bits:

00000000 00000000 00000000 10000000

When the last 8 bits are kept, this is the result:

10000000

Now, the most significant bit in the byte is interpreted as -128, not +128 as it was in the int.

The value 128 overflows a byte, and the result is negative because of that most significant bit being set.

Answer 4

Logic is simple, Java numbers are always signed in two's complement.

Now a byte has 8 bits and 128 is 10000000. When you do

int i = 128

you end up with:

i == 00000000 00000000 00000000 10000000

When you cast it to a byte you the 24 most significative are truncated, so you end up with

b == 10000000

but a Java byte is signed, and 128 can't be represented, since it overflows and wraps around. So what happens is that the value ends up as 128 - 256 = -128 (that's because of two's complement).