1 - 100 0000 1111 - 1000 0001 1100 1101 0110 1110 1001 1110 0001 1011 0000 0100 0000 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

1 - 100 0000 1111 - 1000 0001 1100 1101 0110 1110 1001 1110 0001 1011 0000 0100 0000: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
1 - 100 0000 1111 - 1000 0001 1100 1101 0110 1110 1001 1110 0001 1011 0000 0100 0000, 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.
1

The next 11 bits contain the exponent:
100 0000 1111

The last 52 bits contain the mantissa:
1000 0001 1100 1101 0110 1110 1001 1110 0001 1011 0000 0100 0000


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

The exponent is allways a positive integer.

100 0000 1111(2) =

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

1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 8 + 4 + 2 + 1 =

1,024 + 8 + 4 + 2 + 1 =

1,039(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 = 1,039 - 1023 = 16


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 0001 1100 1101 0110 1110 1001 1110 0001 1011 0000 0100 0000(2) =

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

0.5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0 + 0 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0 + 0.000 000 029 802 322 387 695 312 5 + 0 + 0 + 0.000 000 003 725 290 298 461 914 062 5 + 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 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0 + 0 + 0 + 0 + 0 + 0 =

0.5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 003 725 290 298 461 914 062 5 + 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 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 =

0.507 040 895 080 550 058 082 735 631 614 923 477 172 851 562 5(10)

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

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

(-1)1 × (1 + 0.507 040 895 080 550 058 082 735 631 614 923 477 172 851 562 5) × 216 =

-1.507 040 895 080 550 058 082 735 631 614 923 477 172 851 562 5 × 216 = ...

= -98 765.432 099 998 928 606 510 162 353 515 625

1 - 100 0000 1111 - 1000 0001 1100 1101 0110 1110 1001 1110 0001 1011 0000 0100 0000, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = -98 765.432 099 998 928 606 510 162 353 515 625(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)