1 - 011 1000 0101 - 0001 1110 1011 1000 0111 0000 0000 1101 0000 0000 1010 0011 0011 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

1 - 011 1000 0101 - 0001 1110 1011 1000 0111 0000 0000 1101 0000 0000 1010 0011 0011: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
1 - 011 1000 0101 - 0001 1110 1011 1000 0111 0000 0000 1101 0000 0000 1010 0011 0011, 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:
011 1000 0101

The last 52 bits contain the mantissa:
0001 1110 1011 1000 0111 0000 0000 1101 0000 0000 1010 0011 0011


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

The exponent is allways a positive integer.

011 1000 0101(2) =

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

0 + 512 + 256 + 128 + 0 + 0 + 0 + 0 + 4 + 0 + 1 =

512 + 256 + 128 + 4 + 1 =

901(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 = 901 - 1023 = -122


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).


0001 1110 1011 1000 0111 0000 0000 1101 0000 0000 1010 0011 0011(2) =

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

0 + 0 + 0 + 0.062 5 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0 + 0 + 0 + 0 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0 + 0 + 0 + 0 + 0 + 0 + 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 + 0.000 000 000 232 830 643 653 869 628 906 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 0 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0 + 0 + 0 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0 + 0 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =

0.062 5 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.001 953 125 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 232 830 643 653 869 628 906 25 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =

0.120 001 795 935 092 969 713 764 432 526 659 220 457 077 026 367 187 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 + 0.120 001 795 935 092 969 713 764 432 526 659 220 457 077 026 367 187 5) × 2-122 =

-1.120 001 795 935 092 969 713 764 432 526 659 220 457 077 026 367 187 5 × 2-122 = ...

= -0.000 000 000 000 000 000 000 000 000 000 000 000 210 648 925 445 202 918 131 716 512 163 529 631 260 734 448 000 672 979 641 676 189 320 912 864 382 466 355 504 932 532 932 411 796 966 487 571 580 728 399 567 306 041 717 529 296 875

1 - 011 1000 0101 - 0001 1110 1011 1000 0111 0000 0000 1101 0000 0000 1010 0011 0011, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = -0.000 000 000 000 000 000 000 000 000 000 000 000 210 648 925 445 202 918 131 716 512 163 529 631 260 734 448 000 672 979 641 676 189 320 912 864 382 466 355 504 932 532 932 411 796 966 487 571 580 728 399 567 306 041 717 529 296 875(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)