0 - 011 1111 1100 - 1000 0011 0010 0101 0100 0011 0100 0100 0011 1100 0000 0110 1000 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

0 - 011 1111 1100 - 1000 0011 0010 0101 0100 0011 0100 0100 0011 1100 0000 0110 1000: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
0 - 011 1111 1100 - 1000 0011 0010 0101 0100 0011 0100 0100 0011 1100 0000 0110 1000, a 64 bit double precision IEEE 754 binary floating point representation standard to decimal?

1. Identify the elements that make up the binary representation of the number:

The first bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.
0

The next 11 bits contain the exponent:
011 1111 1100

The last 52 bits contain the mantissa:
1000 0011 0010 0101 0100 0011 0100 0100 0011 1100 0000 0110 1000


2. Convert the exponent from binary (from base 2) to decimal (in base 10).

The exponent is allways a positive integer.

011 1111 1100(2) =

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

0 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 0 + 0 =

512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 =

1,020(10)

3. Adjust the exponent.

Subtract the excess bits: 2(11 - 1) - 1 = 1023,

that is due to the 11 bit excess/bias notation.

The exponent, adjusted = 1,020 - 1023 = -3


4. Convert the mantissa from binary (from base 2) to decimal (in base 10).

The mantissa represents the fractional part of the number (what comes after the whole part of the number, separated from it by a comma).


1000 0011 0010 0101 0100 0011 0100 0100 0011 1100 0000 0110 1000(2) =

1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 1 × 2-7 + 1 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 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 + 1 × 2-23 + 1 × 2-24 + 0 × 2-25 + 1 × 2-26 + 0 × 2-27 + 0 × 2-28 + 0 × 2-29 + 1 × 2-30 + 0 × 2-31 + 0 × 2-32 + 0 × 2-33 + 0 × 2-34 + 1 × 2-35 + 1 × 2-36 + 1 × 2-37 + 1 × 2-38 + 0 × 2-39 + 0 × 2-40 + 0 × 2-41 + 0 × 2-42 + 0 × 2-43 + 0 × 2-44 + 0 × 2-45 + 1 × 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 + 0.007 812 5 + 0.003 906 25 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0 + 0 + 0 + 0 + 0.000 000 119 209 289 550 781 25 + 0.000 000 059 604 644 775 390 625 + 0 + 0.000 000 014 901 161 193 847 656 25 + 0 + 0 + 0 + 0.000 000 000 931 322 574 615 478 515 625 + 0 + 0 + 0 + 0 + 0.000 000 000 029 103 830 456 733 703 613 281 25 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 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 + 0.007 812 5 + 0.003 906 25 + 0.000 488 281 25 + 0.000 061 035 156 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 000 119 209 289 550 781 25 + 0.000 000 059 604 644 775 390 625 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 029 103 830 456 733 703 613 281 25 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 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 =

0.512 287 334 593 588 994 380 297 663 155 943 155 288 696 289 062 5(10)

5. Put all the numbers into expression to calculate the double precision floating point decimal value:

(-1)Sign × (1 + Mantissa) × 2(Adjusted exponent) =

(-1)0 × (1 + 0.512 287 334 593 588 994 380 297 663 155 943 155 288 696 289 062 5) × 2-3 =

1.512 287 334 593 588 994 380 297 663 155 943 155 288 696 289 062 5 × 2-3 = ...

= 0.189 035 916 824 198 624 297 537 207 894 492 894 411 087 036 132 812 5

0 - 011 1111 1100 - 1000 0011 0010 0101 0100 0011 0100 0100 0011 1100 0000 0110 1000, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 0.189 035 916 824 198 624 297 537 207 894 492 894 411 087 036 132 812 5(10)

Spaces were used to group digits: for binary, by 4, for decimal, by 3.

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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)