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Floating Point Numbers

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• Representing decimal point numbers representation
  • A floating point number can be written as 2 numbers:

    An exponent,    and      A mantissa

  • Example:

    314.159 = 0.314159 x 103 = 314159 x 10-3

    (Remember that multiply/dividing by 10 with decimal number can be accomplished by shifting the decimal point one place to right/left)

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  • We do not need to record the fact that the exponent used is 10 (it is assumed)

    Then, we can represent the number 314.159 using a pair of numbers:

    314.159 = (314.159, 0) = (0.314159, 3) = (314159, -3)

    The first number is the mantissa and the second is the exponent.

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• Representing binary floating point numbers
  • Binary floating point numbers can also be written as an exponent and a mantissa, but now using powers of 2 (instead of 10 for decimal numbers):

  • Example:

    1010.1011 = 0.10101011 x 24 = 10101011 x 2-4

    (Similarly, multiply/dividing by 2 with binary number can be accomplished by shifting the decimal point one place to right/left)

  • Again, we do not need to record the fact that the exponent used is 2 (because every number inside the computer is in binary).

    We can represent the number 1010.1011 using a pair of binary numbers:

    1010.1011 = (1010.1011, 0) = (0.10101011, 00000100 (4) ) = (10101011, 11111100 (-4) )

    The first number is the mantissa and the second is the exponent.

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• Standardizing a floating point representation
  • From the previous discussion, we can see that:

    There are many ways to represent the same floating point value.

  • In order to exchange floating point information, computer scientists have established an international standard to represent (= encode) floating point numbers

  • This international standard is the IEEE 754 standard: click here

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• The IEEE 754 Standard for floating point representation
  • IEEE is a standard making organization (Institute of Electric and Electronics Engineers).

  • IEEE has defined a number of floating point number formats (single precision and double precision)

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  • The IEEE Single Precision represention:

    Uses 32 bits (4 bytes) Format: S EEEEEEEE MMMMMMMMMMMMMMMMMMMMMMM Bit: 0 1 8 9 31 S = sign of the mantissa (1 bit) M = mantissa (23 bits) E = exponent (8 bits) The mantissa uses the sign/magnitude representation: (See: click here) S = the sign of the mantissa MMMM...M = the magnitude of the mantissa The mantissa is always a fixed point binary number between 1 and 2: In other words: 1.0 <= M < 2.0 I.e.: The mantissa is always of the form 1.xxxxxxxxxxx        Because the mantissa will always start with 1.xxxxxxx, the leading "1" of the mantissa is omitted (= not stored, but assumed to be present !)         The exponent EEEEEEEE uses the excess 127 encoding scheme for signed numbers that is similar to the 2's complement representation. The excess 127 encoding for 8 bits: Bit pattern Encodes the value ------------------------------------- 00000000 -127 00000001 -126 .... 01111111 0 10000000 1 10000001 2 .... 11111111 128

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  • Example:

    Given the following floating point representation: 01000000101000000000000000000000 = 0 10000001 01000000000000000000000 ^ ^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^ | exponent mantissa (with leading "1." omitted) sign Meaning of each bit according to the IEEE format: S EEEEEEEE MMMMMMMMMMMMMMMMMMMMMMM 0 10000001 01000000000000000000000 Bit: 0 1 8 9 31 Sign of mantissa = 0 (positive mantissa) Mantissa bits = 01000000000000000000000 Actual mantissa: 1.01000000000000000000000 (leading 1. was omitted !!) = 1.01 Exponent = 10000001 (= 129(10) = 127(10) + 2(10)) Value represented by Exponent = 2 (decimal) Therefore: value represented = 1.01 (binary) with exponent 22 = 101 (binary) = 5 (decimal)

    Hands-on experimention: click here

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  • The IEEE Double Precision represention:
    • Uses 64 bits, and it is similar to the single precision format

          S EEEEEEEEEEE FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
          0 1        11 12                                                63
       

    • The first bit is the sign bit, S of the mantissa
    • The next 11 bits are the exponent bits, 'E',
    • The final 52 bits are the mantissa 'M':
    • The mantissa is normalized so that 1.0 <= M < 2.0

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• Example Program: (Demo above code)                                                
  • Prog file: /home/cs255001/demo/asm/1-directives/float.s

  How to run the program:

  To compile:   as255 float               To run:          use EGTAPI

  Use the Variable display window to show the content (= bit pattern) stored in the variable floatCost as binary Signed Int and as Float

  Try it with different values.

  This demo shows that:

  When the programmer write a floating point number in the program, the assembler translates the number into its IEEE 754 representation (and store this binary pattern in memory) !!!

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• Explore further....

  Here is a nice webpage where you can construct floating point representations: click here

  When you explore the above webpage, you must know that there are 2 things that are strange in the IEEE float representation

  1. Because the first bit of the mantissa is always 1 (stop, think, why is that so ? because we made it so: 1.0 <= mantissa < 2.0), it is not stored.

     So if the mantissa bits in a single precision representation are 01010101010101010101010, the actual mantissa bits are 1.01010101010101010101010 (and the mantissa is between 1.0 and 2.0)

  2. The exponent uses a modified 2's complement encoding, called "excess" encoding.

     The single precision exponent uses the "excess 127" encoding which uses 01111111 to represent 0:

  Bit pattern:
      01111100   01111101   01111110   01111111   10000000   10000001   10000010
      ----+----------+----------+----------+----------+----------+----------+-
         -3         -2         -1          0          1          2          3
  Value represented: