Converting Integer to Byte in Java

Question

I have the following code:

    int i =128;
    byte b = (byte) i;
    System.out.println( Integer.toBinaryString(i)); //10000000
    System.out.println( Integer.toBinaryString(b)); //11111111111111111111111110000000

could someone explain why 1's were added to the left when casting from Integer to Byte and how could a byte carry more than 8 bits !?

`Integer#toBinaryString` accepts an `int`, not a `byte`. The `byte` value is widened to an `int`. — Sotirios Delimanolis, Oct 21 '15 at 02:16

jjmontes · Answer 1 · 2015-10-21T10:39:58.020

1

You are calling .toBinaryString on the Integer class, so your number is treated as an Integer in any case.

The reason the second call has so many 1 is because it is a negative number. In Java, bytes are signed, so the maximum positive value is 127. By casting 128 into a byte you are actually representing -128. When you cast that small negative number into an 32 bit signed integer as you were doing, all those 1s appear at the beginning.

edited Oct 21 '15 at 10:39

answered Oct 21 '15 at 02:16

jjmontes

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Thanks you pointed to the problem but actually there is no Byte.toBinaryString(i) :| – Assem-Hafez Oct 21 '15 at 02:24

Converting Integer to Byte in Java

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