Why output are i=11 and j=44

Question

public class Testing2
{
public static void main(String[] args) {
double x = 267.5;
int y = 300;

byte i = (byte)x;
byte j = (byte)y;

System .out .println ("i = "+i+"\nj = "+j);
}
}

Answer 1

Firstly, there are two narrowing conversions from 32-bit integer and 64-bit double-precision FP to an 8-bit byte are taking place. In the case of Double to byte, what's happing is double type loses its precision and becomes an integer type, namely 267. Then, since a byte only capable of holding 8-bits, stuffing a bigger number into it causes an overflow, meaning it's undefined behavior in languages such as C/C++. However, in Java, it simply calculates the fittest number out of 267 to make it small enough for storing into the 8-bit container, either by successive subtractions from 128 ¹ as

267 - 128 = 139
139 - 128 = 11

Or by taking modulo 128 as

267 % 128 = 11

Now, our byte can safely hold 11 without a problem. The same scenario is happening for 300.

__________
¹ The range of a byte type is from -128 to 127. That is, 1-bit for the sign and 7-bit for magnitude, so just to represent the non-negative integers, there are 128 vacancies are available (from 0 to 127).