Please explain why output of
byte c = (byte) (-512);
System.out.print(c);
is 0.
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What is confusing you about it? Do you know what is [range of `byte`](https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html) and [integer overflow](https://en.wikipedia.org/wiki/Integer_overflow)? – Pshemo Jul 20 '15 at 18:24
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possible duplicate of [How does Java handle integer underflows and overflows and how would you check for it?](http://stackoverflow.com/questions/3001836/how-does-java-handle-integer-underflows-and-overflows-and-how-would-you-check-fo) – byako Jul 20 '15 at 18:25
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Yes I do..its -128 to 127 but why is the output 0 – NIKHIL NAGRALE Jul 20 '15 at 18:27
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Try to print results for `-129`,`-130` ... `-512` and observe what happens. – Pshemo Jul 20 '15 at 18:29
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Got it..If it goes out of range the final value will be value%256(modulo) correct? – NIKHIL NAGRALE Jul 20 '15 at 18:42
1 Answers
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The range of byte in Java is -128 to 127. Since -512 is not in the range, the compiler will ask for explicit typecasting. Hence, you have to cast the -512 (an integer) to byte. What happens when the program runs is that, JVM just removes the upper 24 bits to fit -512 in 8 bits.
-512 = 11111111111111110000001000000000 (int)
(byte) -512 = 00000000 (byte) //truncated the upper 24 bits// = 0
