How the byte range is -128 to 127 in java.How it is possible?

Question

1 byte = 8bits

I converted 1111 1111 binary number to decimal .it is giving me 255.

But when i converted 0111 1111 binaru number to decimal .it is giving me 127.

So on what basis the range is declared.Please help me.

Thanks in advance...

byte is signed(both positive and negative) in java.And its from -128 to 127. — Shashank Kadne, May 18 '13 at 06:39

Paul Vargas · Accepted Answer · 2013-05-18T06:58:44.120

4

The number types in Java are signed, meaning they can be negative or positive. The leftmost bit (the most significant bit) is used to represent the sign, where a 1 means negative and 0 means positive.

Byte

Max 01111111  = +127
Min 10000000  = -128

    11111111  = -1

Short

Max 0111111111111111  = +32767
Min 1000000000000000  = -32768

    0000000011111111  = +255

Binary negative numbers are represented in 2's complement form.

edited May 18 '13 at 06:58

answered May 18 '13 at 06:45

Paul Vargas

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if left most bit is for sign then 10000000 values is 0? – PSR May 18 '13 at 07:00
incase of negative number ,left most bit is 1 which does two work i.e tell number is negative and also calculates as 128. see my answer below – M Sach May 18 '13 at 07:03
`10000000 = -128` because of Two's complement – Paul Vargas May 18 '13 at 07:04
How -128 will come.Can u explain pls – PSR May 18 '13 at 07:05
i see in the link they give -128 for 10000000.How it is possible – PSR May 18 '13 at 07:06

score 2 · Answer 2 · answered May 18 '13 at 06:56

One bit is reserved for detrmining whether number is negative or positive .

So for max postive number value will be

 01111111 which gives the int number as 128(leftmost bit 0 represent its a postive number)

        64+32+6+8+4+2+1= 127

for max negativenumber value will ((leftmost bit 1 represent its a negative number))

  10000000 which gives the int number as -128

  -128+0+0+0+0+0+0+0 = -128

so range becomes from

  -127 to 128

How the byte range is -128 to 127 in java.How it is possible?

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