The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 100 0000 1010 - 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1110 Converted and Written as a Base Ten Decimal System Number (Double)

Number 0 - 100 0000 1010 - 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 converted from 64 bit double precision IEEE 754 binary floating point standard representation to decimal system written in base ten (double) = ?

Number 0 - 100 0000 1010 - 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 converted from 64 bit double precision IEEE 754 binary floating point standard representation to decimal system written in base ten (double) = ?

A number in 64 bit double precision IEEE 754 binary floating point standard representation...

... requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Follow the steps below to convert a number from 64 bit double precision IEEE 754 binary floating point representation to base 10 decimal system:

• 1. Identify the elements that make up the binary representation of the number:
  First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
  The next 11 bits contain the exponent.
  The last 52 bits contain the mantissa.
• 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
• 3. Adjust the exponent, subtract the excess bits, 2^{(11 - 1)} - 1 = 1,023, that is due to the 11 bit excess/bias notation.
• 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
• 5. Put all the numbers into expression to calculate the double precision floating point decimal value:
  (-1)^Sign × (1 + Mantissa) × 2^{(Exponent adjusted)}

Example: convert the number 1 - 100 0011 1101 - 1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000 from 64 bit double precision IEEE 754 binary floating point system to base ten decimal (double):

• 1. Identify the elements that make up the binary representation of the number:
  
  First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.

  ---

  The next 11 bits contain the exponent: 100 0011 1101

  ---

  The last 52 bits contain the mantissa:
  1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000
• 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):
  
  100 0011 1101₍₂₎ =

  ---

  1 × 2¹⁰ + 0 × 2⁹ + 0 × 2⁸ + 0 × 2⁷ + 0 × 2⁶ + 1 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰ =

  ---

  1,024 + 0 + 0 + 0 + 0 + 32 + 16 + 8 + 4 + 0 + 1 =

  ---

  1,024 + 32 + 16 + 8 + 4 + 1 =

  ---

  1,085₍₁₀₎
• 3. Adjust the exponent, subtract the excess bits, 2^{(11 - 1)} - 1 = 1,023, that is due to the 11 bit excess/bias notation:
  Exponent adjusted = 1,085 - 1,023 = 62
• 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10):
  
  1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000₍₂₎ =

  ---

  1 × 2^-1 + 0 × 2^-2 + 0 × 2^-3 + 0 × 2^-4 + 0 × 2^-5 + 0 × 2^-6 + 0 × 2^-7 + 0 × 2^-8 + 0 × 2^-9 + 0 × 2^-10 + 1 × 2^-11 + 0 × 2^-12 + 0 × 2^-13 + 0 × 2^-14 + 0 × 2^-15 + 1 × 2^-16 + 0 × 2^-17 + 1 × 2^-18 + 0 × 2^-19 + 0 × 2^-20 + 0 × 2^-21 + 0 × 2^-22 + 0 × 2^-23 + 0 × 2^-24 + 0 × 2^-25 + 1 × 2^-26 + 0 × 2^-27 + 0 × 2^-28 + 1 × 2^-29 + 1 × 2^-30 + 1 × 2^-31 + 0 × 2^-32 + 0 × 2^-33 + 0 × 2^-34 + 0 × 2^-35 + 0 × 2^-36 + 0 × 2^-37 + 1 × 2^-38 + 0 × 2^-39 + 0 × 2^-40 + 0 × 2^-41 + 0 × 2^-42 + 0 × 2^-43 + 0 × 2^-44 + 1 × 2^-45 + 0 × 2^-46 + 1 × 2^-47 + 0 × 2^-48 + 1 × 2^-49 + 0 × 2^-50 + 0 × 2^-51 + 0 × 2^-52 =

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  0.5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 014 901 161 193 847 656 25 + 0 + 0 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0 + 0 + 0 =

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  0.5 + 0.000 488 281 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 =

  ---

  0.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5₍₁₀₎
• 5. Put all the numbers into expression to calculate the double precision floating point decimal value:
  
  (-1)^Sign × (1 + Mantissa) × 2^{(Exponent adjusted)} =

  ---

  (-1)¹ × (1 + 0.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5) × 2⁶² =

  ---

  -1.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5 × 2⁶² =

  ---

  -6 919 868 872 153 800 704₍₁₀₎

• 1 - 100 0011 1101 - 1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000 converted from 64 bit double precision IEEE 754 binary floating point representation to a decimal number (float) in decimal system (in base 10) = -6 919 868 872 153 800 704₍₁₀₎

Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):

1. Integer -> Binary

• 1.1. Unsigned integer -> Unsigned binary

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  1.2. Signed integer -> Signed binary

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  1.3. Signed integer -> Signed binary one's complement

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  1.4. Signed integer -> Signed binary two's complement

2. Decimal -> Binary

• 2.1. Decimal -> 32bit single precision IEEE 754 binary floating point

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  2.2. Decimal -> 64bit double precision IEEE 754 binary floating point

3. Binary -> Integer

• 3.1. Unsigned binary -> Unsigned integer

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  3.2. Signed binary -> Signed integer

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  3.3. Signed binary one's complement -> Signed integer

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  3.4. Signed binary two's complement -> Signed integer

4. Binary -> Decimal

• 4.1. 32bit single precision IEEE 754 binary floating point -> Float

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  4.2. 64bit double precision IEEE 754 binary floating point -> Double