64 bit double precision IEEE 754 binary floating point number 0 - 011 1111 1100 - 0100 1000 1011 0100 0011 1001 0101 1000 0001 0000 0110 0000 0000 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1100 - 0100 1000 1011 0100 0011 1001 0101 1000 0001 0000 0110 0000 0000.

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:
011 1111 1100


The last 52 bits contain the mantissa:
0100 1000 1011 0100 0011 1001 0101 1000 0001 0000 0110 0000 0000

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

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:

Exponent adjusted = 1,020 - 1023 = -3

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):

0100 1000 1011 0100 0011 1001 0101 1000 0001 0000 0110 0000 0000(2) =

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


0 + 0.25 + 0 + 0 + 0.031 25 + 0 + 0 + 0 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0 + 0 + 0.000 000 059 604 644 775 390 625 + 0 + 0.000 000 014 901 161 193 847 656 25 + 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 + 0 + 0 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 0 + 0 + 0 + 0 + 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 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.25 + 0.031 25 + 0.001 953 125 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 061 035 156 25 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0.000 000 059 604 644 775 390 625 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 014 551 915 228 366 851 806 640 625 + 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 =


0.283 999 999 999 991 814 547 684 043 645 858 764 648 437 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 + 0.283 999 999 999 991 814 547 684 043 645 858 764 648 437 5) × 2-3 =


1.283 999 999 999 991 814 547 684 043 645 858 764 648 437 5 × 2-3 =


0.160 499 999 999 998 976 818 460 505 455 732 345 581 054 687 5

0 - 011 1111 1100 - 0100 1000 1011 0100 0011 1001 0101 1000 0001 0000 0110 0000 0000
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =


0.160 499 999 999 998 976 818 460 505 455 732 345 581 054 687 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 - 011 1111 1100 - 0100 1000 1011 0100 0011 1001 0101 1000 0001 0000 0110 0000 0000 = 0.160 499 999 999 998 976 818 460 505 455 732 345 581 054 687 5 Sep 19 02:13 UTC (GMT)
1 - 110 1100 0000 - 1011 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -291 280 025 548 241 182 963 469 814 764 487 124 854 850 066 581 333 252 469 717 305 007 431 043 044 200 850 524 544 284 446 371 460 486 694 181 900 410 868 805 858 451 060 853 928 138 920 702 742 924 847 902 066 270 510 788 737 619 624 863 380 592 844 370 468 897 863 814 868 893 696 Sep 19 02:09 UTC (GMT)
0 - 000 0000 0000 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1101 = 0 Sep 19 02:09 UTC (GMT)
1 - 011 0001 0001 - 0110 0010 0110 1110 0000 1111 1100 0010 1010 0000 1111 1000 0000 = -0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 003 134 378 597 404 520 868 447 628 062 751 383 742 974 798 936 832 529 084 769 679 970 619 375 026 514 144 264 011 097 134 135 042 457 444 470 965 263 177 717 333 528 112 431 939 286 331 557 413 606 107 239 593 356 768 123 262 861 419 631 164 494 603 581 260 889 768 600 463 867 187 5 Sep 19 02:06 UTC (GMT)
0 - 100 0000 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 8 Sep 19 02:06 UTC (GMT)
0 - 011 1111 1111 - 1010 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 = 1.648 437 500 000 000 444 089 209 850 062 616 169 452 667 236 328 125 Sep 19 02:06 UTC (GMT)
0 - 100 0000 0001 - 1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 6.003 906 25 Sep 19 02:05 UTC (GMT)
0 - 100 0010 0001 - 1110 1101 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 = 33 103 460 433.919 998 168 945 312 5 Sep 19 02:04 UTC (GMT)
0 - 111 1111 1111 - 0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 = SNaN, Signalling Not a Number Sep 19 02:03 UTC (GMT)
0 - 100 0001 1100 - 1111 1110 1010 1010 1100 0100 1011 0000 0000 0000 0000 0000 0000 = 1 070 946 454 Sep 19 02:03 UTC (GMT)
0 - 101 1111 0000 - 0011 0010 0111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 489 789 860 550 377 917 134 847 078 965 628 620 706 732 054 191 563 313 685 198 290 337 382 324 333 865 946 734 395 546 346 249 636 307 092 206 185 224 213 505 804 205 592 177 614 811 415 703 453 696 Sep 19 02:01 UTC (GMT)
0 - 100 0000 0111 - 1001 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 410 Sep 19 01:59 UTC (GMT)
0 - 000 0000 0000 - 0110 0000 1110 1001 1110 1010 0111 0110 0100 0001 0000 0000 0000 = 0 Sep 19 01:58 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)