64 bit double precision IEEE 754 binary floating point number 0 - 100 0001 0111 - 1111 0001 0011 0110 0000 0001 1100 1001 0100 1000 1000 0110 1000 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
0 - 100 0001 0111 - 1111 0001 0011 0110 0000 0001 1100 1001 0100 1000 1000 0110 1000.

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

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


The next 11 bits contain the exponent:
100 0001 0111


The last 52 bits contain the mantissa:
1111 0001 0011 0110 0000 0001 1100 1001 0100 1000 1000 0110 1000

2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):

100 0001 0111(2) =


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


1,024 + 0 + 0 + 0 + 0 + 0 + 16 + 0 + 4 + 2 + 1 =


1,024 + 16 + 4 + 2 + 1 =


1,047(10)

3. Adjust the exponent, subtract the excess bits, 2(11 - 1) - 1 = 1023, that is due to the 11 bit excess/bias notation:

Exponent adjusted = 1,047 - 1023 = 24

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):

1111 0001 0011 0110 0000 0001 1100 1001 0100 1000 1000 0110 1000(2) =

1 × 2-1 + 1 × 2-2 + 1 × 2-3 + 1 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 1 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 1 × 2-12 + 0 × 2-13 + 1 × 2-14 + 1 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 + 1 × 2-24 + 1 × 2-25 + 1 × 2-26 + 0 × 2-27 + 0 × 2-28 + 1 × 2-29 + 0 × 2-30 + 0 × 2-31 + 1 × 2-32 + 0 × 2-33 + 1 × 2-34 + 0 × 2-35 + 0 × 2-36 + 1 × 2-37 + 0 × 2-38 + 0 × 2-39 + 0 × 2-40 + 1 × 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 + 0.25 + 0.125 + 0.062 5 + 0 + 0 + 0 + 0.003 906 25 + 0 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 059 604 644 775 390 625 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 014 901 161 193 847 656 25 + 0 + 0 + 0.000 000 001 862 645 149 230 957 031 25 + 0 + 0 + 0.000 000 000 232 830 643 653 869 628 906 25 + 0 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0 + 0 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0 + 0 + 0 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 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 + 0.25 + 0.125 + 0.062 5 + 0.003 906 25 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 000 059 604 644 775 390 625 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 232 830 643 653 869 628 906 25 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 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 =


0.942 230 331 078 940 608 335 869 910 661 131 143 569 946 289 062 5(10)

Conclusion:

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

(-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =


(-1)0 × (1 + 0.942 230 331 078 940 608 335 869 910 661 131 143 569 946 289 062 5) × 224 =


1.942 230 331 078 940 608 335 869 910 661 131 143 569 946 289 062 5 × 224 =


32 585 217.786 262 899 637 222 290 039 062 5

0 - 100 0001 0111 - 1111 0001 0011 0110 0000 0001 1100 1001 0100 1000 1000 0110 1000
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =


32 585 217.786 262 899 637 222 290 039 062 5(10)

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

64 bit double precision IEEE 754 binary floating point standard representation of numbers requires three building blocks: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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

0 - 100 0001 0111 - 1111 0001 0011 0110 0000 0001 1100 1001 0100 1000 1000 0110 1000 = 32 585 217.786 262 899 637 222 290 039 062 5 Aug 13 16:19 UTC (GMT)
0 - 100 0000 0111 - 0010 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 289 Aug 13 16:15 UTC (GMT)
0 - 000 0000 0001 - 0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1000 = 0 Aug 13 16:13 UTC (GMT)
0 - 100 0000 0111 - 0011 1101 0001 1000 0111 1100 0110 0101 0011 0111 1100 1101 1001 = 317.095 648 122 905 515 720 049 152 150 750 160 217 285 156 25 Aug 13 15:57 UTC (GMT)
0 - 100 0001 1001 - 0001 1011 1010 1110 0001 1001 1011 1011 0100 0101 1110 1110 0011 = 74 365 030.926 143 214 106 559 753 417 968 75 Aug 13 15:56 UTC (GMT)
0 - 011 1111 1000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0110 = 0.007 812 500 000 000 010 408 340 855 860 842 566 471 546 888 351 440 429 687 5 Aug 13 15:48 UTC (GMT)
0 - 010 1100 0001 - 0010 0000 0000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 = 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 002 106 761 503 282 684 761 438 771 084 883 581 364 842 757 076 894 850 134 843 352 548 773 692 647 207 274 259 382 708 765 442 805 711 124 559 513 178 228 717 425 305 353 217 345 233 316 567 715 692 977 721 829 948 597 721 290 208 953 623 667 106 589 479 318 078 648 485 690 340 663 6 Aug 13 15:48 UTC (GMT)
0 - 100 0000 0111 - 1011 0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 437.5 Aug 13 15:47 UTC (GMT)
0 - 000 0000 0000 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1101 = 0 Aug 13 15:45 UTC (GMT)
0 - 011 1111 1111 - 0011 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1100 = 1.245 000 000 000 000 106 581 410 364 015 027 880 668 640 136 718 75 Aug 13 15:45 UTC (GMT)
0 - 100 0000 0011 - 1110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 = 30.000 000 000 000 056 843 418 860 808 014 869 689 941 406 25 Aug 13 15:45 UTC (GMT)
0 - 100 0010 1101 - 0110 1011 1100 1100 0100 0001 1110 1001 0000 0000 0000 0000 0000 = 100 000 000 000 000 Aug 13 15:23 UTC (GMT)
0 - 110 0101 0110 - 1100 0110 1001 0110 0001 0111 0011 0010 0000 0011 0000 0011 0001 = 3 684 204 870 510 636 558 931 979 970 881 447 794 220 096 953 796 727 533 268 637 489 160 945 120 033 697 995 116 093 422 234 431 107 393 634 820 732 689 918 921 340 846 997 331 637 005 419 926 392 810 872 188 403 887 425 705 830 716 014 592 Aug 13 15:21 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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)