Java Quick Guide

Java double type.

• signbit (bitnumber 63, total 1 bit)
  The sign is 1 (positive number) or -1 (negative number).
  The sign can also be written as sign=(-1)^signbit
  The signbit is 0 (positive number) and 1 (negative number)
  
  
• radix
  For a Java float or double the radix is always 2
  
  
• exponent (bitnumber 52-62, total 11 bits)
  
  e 

  e 

  e 

  e 

  e 

  e 

  e 

  e 

  e 

  e 

  e 
  
  In this tutorial this binary representation is called the exponent field and its unsigned value exponent_field_val.
  
  The exponent indicates the positive or negative power of the radix that the mantissa and sign should be multiplied by.
  The exponent field contains the binary numbers which represents only positive numbers (= unsigned numbers).
  However the exponent needs to represent both positive and negative numbers (= signed numbers). To get signed numbers you must subtract the exponent_field_val by an expbias value.
  The expbias = (2^n-1)-1 = 1023 (n is the number of bits used for the exponent field, which is 11)
  Thus: exponent = exponent_field_val - 1023
  
  The minimum and maximum allowed exponent values are:
  exponent_min = -1022
  exponent_max = 1023
  
  IEEE reserves exponent field values of all 0s (exponent_min - 1 = -1023) and all 1s (exponent_max + 1 = 1024) to denote special values in the floating-point scheme. This will be discussed in Table A.
  
  Example 1a:
  
  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  1 

  1 
  
  The exponent_field_val represent unsigned value 3
  exponent = exponent_field_val - 1023 = 3-1023 = -1020
  
  Example 1b:
  
  1 

  1 

  1 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 
  
  The exponent_field_val represent unsigned value 1792
  exponent = exponent_field_val - 1023 = 1792-1023 = 769
  
  
• mantissa (bitnumber 0-51, total 52 bits)
  The mantissa is composed of the fraction (1.02267) and an implicit leading digit (1.02267).
  
  Show Java double binary representation B.
  
  The fraction is represented as a binary decimal.
  In a binary decimal the highest order bit (bitnumber 51) corresponds to a value of 2^-1 = 0.5
  In a binary decimal the lowest order bit (bitnumber 0) corresponds to a value of 2^-52 = 2.22044....e-16
  
  As displayed above the mantissa of a double has only 52 bits, however there is one hidden bit (bitnumber 52). This hidden bit determines what the "implicit leading digit" will be as mentioned earlier. This hidden bit is predictable and is therefore not included, because the exponent value determines what the hidden bit value will be.
  
  If the exponent is all zeros,
  
  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 
  
  the hidden bit (bitnumber 52) = 0 in all other cases the hidden bit = 1.
  
  Example 2a:
  The exponent is:
  
  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 
  
  Because the exponent is all zeros, the hidden bit is 0.
  
  Show Java double binary representation C.
  
  mantissa = hidden bit value + fraction
  mantissa = 0 + 2^-1 + 2^-2 + 2^-3 = 0 + 0.875 = 0.875
  
  Example 2b:
  The exponent is:
  
  1 

  0 

  1 

  0 

  0 

  0 

  0 

  0 

  0 

  0 

  0 
  
  Because the exponent in not all zeros, the hidden bit is 1.
  
  Show Java double binary representation D.
  
  mantissa = hidden bit value + fraction
  mantissa = 1 + 2^-3 + 2^-4 + 2^-6 + 2^-9 = 1 + 0.205078125 = 1.205078125
  
  Note:
  A floating point number is normalized if the mantissa is within the range 1/radix <= mantissa <1, else it is called denormalized.