Operators and Assignments - Binary/Octal/Hex and Decimal Number Systems

Probably not directly required on exam but helpful when using bitwise and logical operators.

Decimal system

• the decimal number system we use every day is built on base ten 10₁₀
• it is based on 10 positions numbered 0 thru 9
• each position corresponds to a power of 10

    1024 = 1 x 103  -> 1 x 1000  = 1000
           0 x 102  -> 0 x  100  =  000
           2 x 101  -> 2 x   10  =   20
           4 x 100  -> 4 x    1  =    4 
                                   ----
                                   1024

Binary system

• computer memory is based on the electrical representation of data
• each memory position is represented by a bit which can be either 'on' or 'off'. This makes it easier to represent computer memory using a base 2 number system rather than the base 10 decimal system.
• the binary system represents numbers by a series of 1's and 0's which correspond to 'on' and 'off' values
• a 1 represents an 'on' position, a 0, an 'off' position
• a byte is represented by 8 bits numbered 0 to 7 from left to right
• the leftmost bit is called the high-order bit, the right most bit, the low-order bit
• in the decimal system, each position corresponds to a power of 10, in the binary system, each position corresponds to a power of 2

01001001 = 0 x 27    -> 0 x 128  =  0
           1 x 26    -> 1 x  64  = 64
           0 x 25    -> 0 x  32  =  0
           0 x 24    -> 0 x  16  =  0
           1 x 23    -> 1 x   8  =  8
           0 x 22    -> 0 x   4  =  0
           0 x 21    -> 0 x   2  =  0
           1 x 20    -> 1 x   1  =  1
                                   --
                                   73

• the largest number which can be represented by a byte is 255 or
  128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 or the bit pattern: 1111 1111
• the smallest number is 0 represented by the bit pattern: 0000 0000
• 0 to 255 gives 256 possible values

Two's-complement

• the two's complement method allows us to represent negative and positive values within the 0 to 256 bit positions
• in this system the numbers 0 thru 127 represent themselves and the numbers 128 to 256 represent negative numbers where 255 = -1, 254 = -2, 253 = -3, ...
• -1 is represented by 256 - 1 = 255, -127 is represented by 256-127 = 129, and -50 would be represented by 256 - 50 = 206
• the high-order bit (the 7th position) is reserved for the sign value of a number
• a 0 in the high order bit means 'the sign value is set to positive'
• a 1 in the high-order bit means 'the sign value is set to negative'

01111111 =   0 +  64 + 32 + 16 + 8 + 4 + 2 + 1  =   127
10000000 = 128 or set sign negative
11111111 =   0 -  64 - 32 - 16 - 8 - 4 - 2 - 1  =  -127  

• larger numbers are represented by increasing the number of bits in a memory block
• this is done in multiples of 8 hence 16-bit, 32-bit and 64-bit memory
• 16-bit memory can store numbers up to 2¹⁶-1, 32-bit, 2³²-1, 64-bit, 2⁶⁴-1
• if signed numbers are being used, the left-most bit still represents the sign
• so, 16-bits allows us to store 0 to 65,535 positions (2¹⁶ - 1), 32-bits, 0 to 4,294,967,295 (2³² - 1)
• using two's-complement arithmetic with 32-bit memory, subtract the negative number from 65,536 to find it's positive complement ie -336 would be represented by 65536 - 336 = 65200

Octal system

• uses base 8
• octal digits are represented by 0 thru 7
• each position is a power of 8
• each octal number can be represented by 3 binary digits
• 2²+2¹+2⁰ = 4+2+1 = 7

• to convert from Octal to Binary just replace the octal digit with the corresponding binary pattern

      Octal: 17       Binary: 001 111
  

• to convert from Binary to Octal just replace the binary pattern with the corresponding octal digit

      Binary: 111 010 Octal:  72
  

Hexidecimal system

• the hexidecimal system uses a base of 16
• hexidecimal digits are represented by 0 thru 9 and the letters A,B,C,D,E,F
• one hexidecimal digit corresponds to a four-digit binary number
• 2³+2²+2¹+2⁰ = 8 + 4 + 2 + 1 = 15

• this makes it easy to convert a number from binary to hex (just replace the binary pattern with the hex digits) or from hex to binary (replace the hex digit with the binary pattern)

Binary: 0000 1111                -> Hex:     0x0F
Binary: 1011 0011 0000 0010      -> Hex:     0xB302

Hex:    0xA0FF                   -> Binary:  1010 0000 1111 1111
Hex:    0xF075                   -> Binary:  1111 0000 0111 0101

Converting between number systems

• to convert a decimal number to a Hex, Octal or Binary number divide by the required base, the resulting remainders, in reverse order represent the required value

Convert Decimal 49 to Binary 49:

    49 / 2 = 24      remainder: 1      ( 49 - 2*24 = 49 - 48 = 1)
    24 / 2 = 12      remainder: 0      ( 24 - 2*12 = 24 - 24 = 0)
    12 / 2 =  6      remainder: 0      ( 12 - 2* 6 = 12 - 12 = 0)
     6 / 2 =  3      remainder: 0      (  6 - 2* 3 =  6 -  6 = 0)
     3 / 2 =  1      remainder: 1      (  3 - 2* 1 =  3 -  2 = 1)
     1 / 2 =  0      remainder: 1      (  1 - 2* 0 =  1 -  0 = 1)
 
 Proof: 110001 = 32+16+0+0+0+0+1 = 4910
 
Convert Decimal 49 to Octal 49:

    49 / 8 = 6      remainder:  1    ( 49 - 8*6 = 49 - 48 = 1)
     6 / 8 = 0      remainder:  6    (  6 - 8*0 =  6 -  0 = 6)
     
 Proof: 618 = (6 * 81) + (1 * 80) = 48 + 1 = 4910     
 
Convert Decimal 49 to Hexidecimal 49:

    49 / 16 = 3     remainder:  1     ( 49 - 16*3 = 49 - 48 = 1)
     3 / 16 = 0     remainder:  3     (  3 - 16*0 =  3 -  0 = 3) 
     
 Proof: 3116 = (3*161) + (1*160) = 48 + 1 = 4910

• in Java octal numbers are represented by 3 digits beginning with a zero, so 61₈ would be written as 061
• Hex numbers are always represented by 4 digits preceeded by 0x, so 31₁₆ would be written as 0x0031
• when converting large decimal numbers to binary, the simplest method is to convert to Hex and then to binary

Convert 4823 to Hex:
    
4823 / 16 = 301 remainder:  7  (4823 - 16*301 = 4823 - 4816 =  7)
 301 / 16 =  18 remainder: 13  ( 301 - 16* 18 =  301 -  288 = 13)
  18 / 16 =   1 remainder:  2  (  18 - 16*  1 =   18 -   16 =  2)
   1 / 16 =   0 remainder:  1  (   1 - 16*  0 =    1 -    0 =  1)
       
 Hex value: 0x12D7 
 Hex value converted to binary: 0001 0010 1101 0111        

• when converting large binary numbers, the simplest method is to convert to Hex and then to decimal

Convert Binary 0001 0010 1101 0111 to decimal

    0001    1   1 * 163 = 4096
    0010    2   2 * 162 =  512
    1101    D  13 * 161 =  208
    0111    7   7 * 160 =    7
                          ----
                          4823

Study aids

• If you have Windows 95 you can use the Calculator in the Scientific mode (Start->Programs->Accessories->Calculator) to check results of decimal to hex, binary, and octal conversions.
• You can also use the Java Integer wrapper class to output binary, hex and octal strings. Example: System.out.println(Integer.toBinaryString(-29));
• Marcus Greene has a great applet that lets you play around with bit-shifting at http://www.software.u-net.com/applets/BitShift/BitShiftAr.html

References:

• C: Step-by-Step by Mitchell Waite and Stephen Prata, SAMS, 1991
  
• Hexidecimal and Octal Notation