Java Bytes

[Richard West]

Hi everyone,

Consider the below program

public class ByteTest
{
 
public void ByteTest1()
{
byte byte1 = (byte)0x9C;
Byte byte2 = new Byte(byte1);
 
System.out.println(byte2.toString());
}
 
public void ByteTest2()
{
byte byte3 = 0x5A;
 
Byte byte4 = new Byte(byte3);
 
System.out.println(byte4.toString());
}
 
public static void main(String[] args) 
{
ByteTest a = new ByteTest();
a.ByteTest1();
a.ByteTest2();
}
 
}

Now from what i understand 1 byte is 8 bits so the maximum is all 1's (ie. 255).

The above first function ByteTest1 i cast the hexadecimal a byte as so

byte byte1 = (byte)0x9C;
Byte byte2 = new Byte(byte1);

Now why do i have to cast it even when 0x9C is well below 255 but if i do not cast it the compiler complains saying that it is an integer.

The above second function ByteTest2 i did not cast the hexadecimal a byte as so

byte byte3 = 0x5A;
Byte byte4 = new Byte(byte3);

Now why is it that the compiler accepts the hexadecimal value 0x5A without any casting but for 0x9C the compiler emit errors saying it is an integer?

Am i missing something?

Any help is greatly appreciated

Thank You

Yours Sincerely

Richard West

[Seb Mathe]

Your missing that Byte are signed : (8 bits, all 1 is -1 !)

So byte range is -128 -> 127

[Layne Lund]

In other words, the maximum value you can store in a byte is 127, NOT 255.

Layne

[Richard West]

Hi everyone,

In that case let's say if i have the number -8 and +10.
How do i convert them to signed bytes respectively?

Richard West
[ October 14, 2005: Message edited by: West Richard ]

[Ernest Friedman-Hill]

Richard,

Please check your private messages.

[Mark Van Tuyl]

int num1 = -8;
int num2 = +10;
 
byte b1 = (byte)num1;
byte b2 = (byte)num2;

[Layne Lund]

Originally posted by West Richard:
Hi everyone,

In that case let's say if i have the number -8 and +10.
How do i convert them to signed bytes respectively?

Richard West

[ October 14, 2005: Message edited by: West Richard ]

Even better than the above, you can just do this:

byte a = -8;
byte b = 10;

This works because these numbers are in the range of valid values for bytes. Also, they are already *signed*.

Perhaps you meant "unsigned" instead. If so, assigning these values to variables is the same, assuming you are actually interested in the bit pattern of the value -8, for instance. Once you do the assignment, you just treat the variable as if its unsigned. You can do any bit-wise operations you wish, as long as you cast the result back to a byte. (The result of &, |, ^, >>, and << is promoted to an int for some reason.)

I hope this answers your question. If not, perhaps you can illustrate what you are trying to do with some more specific code.

Layne