CODEWITHSUNDEEP
javabit-shiftbit
user3383062
user3383062

Reputation:

Shifted bits values

String message = "1";
        byte[] bytes = message.getBytes();

        System.out.println(bytes[0] + ": ");
        for (int i = 0; i < 8; i++) {
            System.out.print((bytes[0] >> (7 - i)) + " ");
        }

Output: 49: 
0 0 1 3 6 12 24 49

So my string is 1 which in ASCII is 49. What I'm trying to understand is why do my bits have values 3,6,12,24 and 49? What's happening behind, why aren't they only 0 and 1 like the first 3?

Upvotes: 0

Views: 68

Answers (5)

JDGuide
JDGuide

Reputation: 6525

Your actual data might be 49 but, it needs to fill 8 bit for byte data types.So, if you count 8 bit starting from 0 to 7 (as per your loop).And you are using >> which will right shift.

49 binary is  0 0 1 1 0 0 0 1

0 0 1 1 0 0 0 1  >> 7-0 = 00000000 = 0
0 0 1 1 0 0 0 1  >> 7-1 = 00000000 = 0
0 0 1 1 0 0 0 1  >> 7-2 = 00000001 = 1
0 0 1 1 0 0 0 1  >> 7-3 = 00000011 = 3
0 0 1 1 0 0 0 1  >> 7-4 = 00000110 = 6
0 0 1 1 0 0 0 1  >> 7-5 = 00001100 = 12
0 0 1 1 0 0 0 1  >> 7-6 = 00011000 = 32
0 0 1 1 0 0 0 1  >> 7-7 = 00110001 = 49

Upvotes: 0

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477794

When shifting, you shift the bits n positions (in this case to the right).

So:

Loop#    7-i     bits      result
0        7      000000000   0
1        6      000000000   0
2        5      000000001   1
3        4      000000011   3
4        3      000000110   6
5        2      000001100  12
6        1      000011000  24
7        0      000110001  49

The reason why the first shifts are 0 and 1 is because al significant bits were already shifted out.

If you want to obtain the last bit, you need to perform (a>>s)&1 with a the number and s the bit from right you want.

In case you want to print the binary representation of a, you can simply use Integer.toBinaryString(a);

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 727077

The last 8 bits of number 49 in binary looks like this:

00110001

When you shift the number right by k bits, it's the same as dividing it in int by 2k. That is what you get in the output (digits to the right of | are dropped):

0 | 0110001 -- 0
00 | 110001 -- 0
001 | 10001 -- 1
0011 | 0001 -- 3
00110 | 001 -- 6
001100 | 01 -- 12
0011000 | 1 -- 24
00110001 |  -- 49

Upvotes: 1

rgettman
rgettman

Reputation: 178343

Because your bit extraction is incorrect. The bit representation for the character '1' is that of 49: 00110001.

You are shifting 7 times, then 6, then 5, etc., but you are not isolating the bits properly.

00110001 >> 7 is 00000000 or 0
00110001 >> 6 is 00000000 or 0
00110001 >> 5 is 00000001 or 1
00110001 >> 4 is 00000011 or 3
00110001 >> 3 is 00000110 or 6
00110001 >> 2 is 00001100 or 12
00110001 >> 1 is 00011000 or 24
00110001 >> 0 is 00110001 or 49

You must do a bitwise-and with 1 to isolate the bit you've shifted to get the 1s and 0s out.

System.out.print( ((bytes[0] >> (7 - i)) & 1) + " ");

Output:

49: 
0 0 1 1 0 0 0 1

Upvotes: 2

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 280181

49 in binary is 

110001

You shift this same value by 7, 6, 5, 4, ..., (7 - i) bits.

So 

00110001 >> 7 ==> 00000000 == 0
00110001 >> 6 ==> 00000000 == 0
00110001 >> 5 ==> 00000001 == 1
00110001 >> 4 ==> 00000011 == 3
...

You can use Integer.toBinaryString(int) to get the binary representation of an integer value as a String.

Upvotes: 4

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