java byte to binary conversion

Question

Based on my statements below I have few questions. Please explain.

Q1. Why the output is same for items 2 and 4. I guess it is because byte 128 is equal to byte -128

Q2. Why the output for items 2 and 4 are padded with 1s on the left where the output is supposed to be just this 10000000. I guess.

Q3. What is the difference between outputs 2 and 3 even though the last out 8 bits looks same.

1. System.out.println("1==>"+Integer.toBinaryString((byte)127));
2. System.out.println("2==>"+Integer.toBinaryString((byte)128));
3. System.out.println("3==>"+Integer.toBinaryString(128));
4. System.out.println("4==>"+Integer.toBinaryString((byte)-128));

output :

1==>1111111

2==>11111111111111111111111110000000

3==>10000000

4==>11111111111111111111111110000000

Answer 1

A byte in Java is actually a signed 8-bit integer. It can only represent numbers from -128 to 127.

Q1. 128 is an overflow. It becomes -128.

Q2. Both these are negative numbers. For negative numbers, the method returns "the argument plus 2^32". This results in a lot of leading 1s. For positive numbers, the output just omits the leading 0.

Q3. In example 3, you don't use a byte. You use int. This means you can represent 128, it is not an overflow.