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two's complement

 
suresh kamsa
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1. byte b1 = (byte)(2 * Byte.MAX_VALUE);
I think b1 = 0111 1111 1111 1111 1111 1111 1111 1111
2.byte b2 = -2;
for negtive number
1000 0000 0000 0000 0000 0000 0000 0000
binary of 2 is 10
which is equal to
1000 0000 0000 0000 0000 0000 0000 0010 if I invert also because of negtive we should get
0111 1111 1111 1111 1111 1111 1111 1101
How is (b1 == b2).Did I do anything wrong?
Thanks.
 
Kavitha Chandrasekaran
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Hi,
b1 will be 11111110 and because it is a byte, it has 8 bits with the most significant bit as 1(that makes it negative). When you do the conversion to decimal,the value is -2.
So, b1 == b2 is true.
Kavitha
 
Ron Newman
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A byte is only 8 bits, not 32 as you are showing here.
 
suresh kamsa
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1. byte b1 = (byte)(2 * Byte.MAX_VALUE);
2. byte b2 = -2;
3. System.out.println(b1 == b2);
as per my knowledge MAX_VALUE = 0111 ... 1111
according to that b1 = 0111 1111(most significant seven bits set to 1 and least one bit set to zero)
MIN_VALUE = 1000 ... 0000
how b1 becomes -2 here, please explain me in detail.
 
Valentin Crettaz
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As you said Byte.MAX_VALUE is 01111111. Now, when you multiply a number by 2, what happens is a shift left by one bit. So,
Byte.MAX_VALUE * 2
can be resolved to
01111111 << 1
which gives
11111110
whose value is -2.
To be sure, let's take the byte value 2 and make it negative (keyword: 2's complement):
2 is 00000010
to get -2 we invert all bits like this
11111101
and add 1, that is
11111110
As you can see we get the same value...
So Byte.MAX_VALUE * 2 is equal to -2
 
suresh kamsa
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Thanks for your in-detail explanation.
Is the same rule applies wheh I multiply number by 3 or 4 etc.(shift left by one bit)
eg.(Byte.MAX_VALUE * 3)
Is the same rule applies to Short.MAXVALUE,Integer.MAXVALUE etc. If anybody has a chart kind which shows this please pass on to me.
What about Byte.MINVALUE,Short.MINVALUE,Integer.MINVALUE etc
Is this possible only for Wrapper primitives?
Thanks
 
It is sorta covered in the JavaRanch Style Guide.
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