public class Test{
public static void main(String[] args) {
byte i = 31;
System.out.println(i<<3);
}
}
Why does this code print 248,not -8?
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Bean Jelly
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5This code prints `256` just as it should. Where are you getting `248` and `-8`? – Ben May 15 '18 at 14:44
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Sorry. It's my mistake. The value of variable i should be 31 – Bean Jelly May 15 '18 at 14:57
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Then you should [edit] your question to fix it. – azurefrog May 15 '18 at 14:58
2 Answers
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The output is 256 and it is ok! You shift left it for 3. i is 32, it means 100000 as binary so if you shift left it for 3, it will be 100000000 that means 256 as dec.
Majva
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I think the data type of (i<<3) is byte,so it means 11111000 as binary ,it should be -8. Am i miss something? – Bean Jelly May 15 '18 at 15:11
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Input is a byte, but when the operation is performed, it is being converted to integer. – Imaskar May 15 '18 at 15:13
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After you now edited your question:
byte b = 31 is 0b11111. If you now do i << 3 this turns to 0b11111000.
Which is 248.
You are expecting -8 which would be the result of (byte) (i << 3). The result of a bitshift is an integer. You can convert that explicitely to a byte if you want a byte-value.
Ben
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Thank you very much! I did not know the result of a bitshift is an integer – Bean Jelly May 15 '18 at 15:24
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1See [JLS 4.2.2](https://docs.oracle.com/javase/specs/jls/se8/html/jls-4.html#jls-4.2.2) "The numerical operators, which result in a value of type int or long" - **all** numeric operators promote byte, short or char to int (or long, in some circumstances). Including bitshift operators. – Erwin Bolwidt May 16 '18 at 13:31