64 Bit IEEE 754 Binary to Double: Converter of 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation to Base Ten Decimal System Double Numbers. Steps and Explanations

Convert 64 bit double precision IEEE 754 binary floating point standard numbers to base ten decimal system (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)

The latest 64 bit double precision IEEE 754 floating point binary standard representation numbers converted and written as decimal system numbers (in base ten, double)

The number 0 - 100 0000 0000 - 1110 0000 0000 0000 0111 1111 1111 1111 1111 1111 1111 1010 1010 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:20 UTC (GMT)
The number 1 - 100 0100 0000 - 1001 1011 1000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0111 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:20 UTC (GMT)
The number 1 - 100 0100 0110 - 1001 0011 1100 0000 0000 0000 0000 0000 0000 0000 0000 0011 1010 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:12 UTC (GMT)
The number 0 - 100 1000 0000 - 0000 1110 0100 1001 1100 0000 0000 0000 0000 0011 0010 1011 1111 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:10 UTC (GMT)
The number 0 - 011 1110 0010 - 0101 0111 1001 1000 1110 1110 0010 0011 0000 1000 1011 1011 0101 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:10 UTC (GMT)
The number 0 - 100 0000 1000 - 0011 0101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 1010 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:10 UTC (GMT)
The number 0 - 100 0001 1011 - 0101 0001 1000 1111 1011 0101 1111 0111 1011 1001 0010 1000 0001 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:05 UTC (GMT)
The number 1 - 100 0000 1101 - 1000 0011 1101 0100 1101 0001 1011 0110 1100 0010 1110 1111 0001 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 20:05 UTC (GMT)
The number 0 - 001 0001 1000 - 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0101 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 19:59 UTC (GMT)
The number 0 - 100 0001 1101 - 1000 1000 0010 1001 1011 1101 1111 1000 1111 1101 1101 0011 0101 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Dec 14 19:48 UTC (GMT)
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How to convert numbers from 64 bit double precision IEEE 754 binary floating point standard to decimal system in base 10

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 =
    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 =
    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)