0 - 111 1111 0000 - 1010 1111 0010 1011 0000 0100 0001 1000 1001 0011 0111 0001 0000 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

0 - 111 1111 0000 - 1010 1111 0010 1011 0000 0100 0001 1000 1001 0011 0111 0001 0000: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
0 - 111 1111 0000 - 1010 1111 0010 1011 0000 0100 0001 1000 1001 0011 0111 0001 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.
0

The next 11 bits contain the exponent:
111 1111 0000

The last 52 bits contain the mantissa:
1010 1111 0010 1011 0000 0100 0001 1000 1001 0011 0111 0001 0000


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

The exponent is allways a positive integer.

111 1111 0000(2) =

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

1,024 + 512 + 256 + 128 + 64 + 32 + 16 + 0 + 0 + 0 + 0 =

1,024 + 512 + 256 + 128 + 64 + 32 + 16 =

2,032(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 = 2,032 - 1023 = 1009


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).


1010 1111 0010 1011 0000 0100 0001 1000 1001 0011 0111 0001 0000(2) =

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

0.5 + 0 + 0.125 + 0 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0 + 0 + 0.000 488 281 25 + 0 + 0.000 122 070 312 5 + 0 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0 + 0 + 0 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 0 + 0 + 0 + 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 + 0 + 0 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0 + 0 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 0 + 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 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 056 843 418 860 808 014 869 689 941 406 25 + 0 + 0 + 0 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0 + 0 + 0 + 0 =

0.5 + 0.125 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.000 488 281 25 + 0.000 122 070 312 5 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 000 238 418 579 101 562 5 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 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 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 056 843 418 860 808 014 869 689 941 406 25 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 =

0.684 250 122 070 299 227 061 696 001 328 527 927 398 681 640 625(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 + 0.684 250 122 070 299 227 061 696 001 328 527 927 398 681 640 625) × 21009 =

1.684 250 122 070 299 227 061 696 001 328 527 927 398 681 640 625 × 21009 = ...

= 9 240 005 132 558 576 839 204 593 043 034 584 475 150 725 194 539 978 737 877 564 051 505 102 414 701 087 819 195 406 323 457 268 171 877 313 858 016 324 746 198 828 505 827 224 809 981 784 241 138 972 652 590 065 566 634 434 963 239 851 530 304 681 748 502 860 358 043 312 405 377 498 284 326 658 907 968 655 535 852 401 625 212 138 487 164 662 371 022 833 236 173 102 923 030 186 113 448 344 723 390 464

0 - 111 1111 0000 - 1010 1111 0010 1011 0000 0100 0001 1000 1001 0011 0111 0001 0000, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 9 240 005 132 558 576 839 204 593 043 034 584 475 150 725 194 539 978 737 877 564 051 505 102 414 701 087 819 195 406 323 457 268 171 877 313 858 016 324 746 198 828 505 827 224 809 981 784 241 138 972 652 590 065 566 634 434 963 239 851 530 304 681 748 502 860 358 043 312 405 377 498 284 326 658 907 968 655 535 852 401 625 212 138 487 164 662 371 022 833 236 173 102 923 030 186 113 448 344 723 390 464(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)