CODEWITHSUNDEEP
javabyte
stackoverflow
stackoverflow

Reputation: 19474

byte value larger than 127

//Ok makes sense

Byte b = (byte)207;
System.out.println(b); //value = 207

//ok doesn't make sense

Integer x = Integer.parseInt("11001111", 2); //207
Byte sens = (byte)x.intValue(); //207
System.out.println(sens); //Value = -49
System.out.println(sens.intValue()); //Value = -49

Whats going on here?

How do I declare/represent an 8 bit byte with a value higher than 127 then

Upvotes: 3

Views: 9685

Answers (4)

Pica
Pica

Reputation: 405

255 + the negative byteValue + 1 returns your number with the counter operation. Still recommend storage in a int.

If you like it bitwise, better put a numeric ring around it.

Upvotes: 0

DJREF
DJREF

Reputation: 1

The reason why it can't go over 127 positive, but it can go 128 negative, is that the first 1 out of the 8 digits in a byte represents if it is a negative number or positive:

1 = negative

0 = positive

and since you can use "1111 1111" for negative numbers, you can get -127.

Upvotes: 0

Marcin Szymczak
Marcin Szymczak

Reputation: 11443

Since Java's Byte is signed you can't represent value larger than 127 in byte.

In Your example:

Byte b = (byte)207;
System.out.println(b); //value = 207

There is an error. Output of println is -49.

Byte b = (byte)207;
System.out.println(b); //value = -49

Which means that both cases are identical.

Upvotes: 1

Peter Elliott
Peter Elliott

Reputation: 3322

bytes in Java are signed, so they go from -128 to 127. Casting an int like that will pick up the high bit at 1 (indicating a negative number in two's complement signed numbers) and convert it to the negative number -49.

From there, when you convert it back to an integer with sens.intValue(), it picks up the new negative value and returns it, so you still get -49.

You will need to store bytes larger than 128 in an int datatype, unfortunately.

Upvotes: 9

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