Converter of 64 bit double precision IEEE 754 binary floating point standard system numbers: converting to base ten decimal (double)

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

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: 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 0000 0110 - 0110 1110 0110 0000 0000 0000 0111 0011 0101 1010 1010 0101 0110 = 183.187 503 437 819 657 392 537 919 804 453 849 792 480 468 75 May 20 15:21 UTC (GMT)
0 - 100 0001 0111 - 0010 1110 0111 1111 1101 1000 0000 1111 0100 0000 0000 1001 0110 = 19 824 600.059 570 871 293 544 769 287 109 375 May 20 15:02 UTC (GMT)
0 - 100 0000 0100 - 1011 1011 1111 1111 1111 1101 1000 0000 0000 0000 0000 0000 0000 = 55.499 995 231 628 417 968 75 May 20 15:02 UTC (GMT)
0 - 011 1110 1100 - 1111 1111 0101 0011 1000 1111 0100 0000 0100 0000 0100 0000 0000 = 0.000 003 809 678 588 706 944 674 102 672 365 734 179 038 554 430 007 934 570 312 5 May 20 14:58 UTC (GMT)
0 - 100 0000 0011 - 0111 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 23.625 May 20 14:54 UTC (GMT)
1 - 100 0011 0101 - 0110 1010 1010 1111 1010 1010 0010 0101 0111 0100 0101 0101 0110 = -25 521 771 719 234 904 May 20 14:52 UTC (GMT)
0 - 100 0000 0110 - 1001 0000 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 200.312 5 May 20 14:47 UTC (GMT)
0 - 011 1111 1110 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 = 0.500 000 000 000 000 111 022 302 462 515 654 042 363 166 809 082 031 25 May 20 14:41 UTC (GMT)
1 - 110 1100 0011 - 1010 1110 0101 0000 0110 1011 0111 1000 0101 1000 1111 1110 1001 = -2 263 510 856 040 996 890 909 801 068 763 326 089 028 066 175 995 255 913 456 035 973 820 496 488 671 254 857 797 234 006 032 254 568 777 973 261 002 270 304 016 658 858 056 665 537 667 789 867 898 119 566 812 543 029 850 169 515 689 129 266 333 676 802 108 436 854 718 131 651 739 648 May 20 14:33 UTC (GMT)
0 - 100 0000 0001 - 0110 1010 1011 1011 1001 1000 1100 0111 1110 0010 1000 0000 0000 = 5.667 699 999 999 967 985 786 497 592 926 025 390 625 May 20 14:33 UTC (GMT)
1 - 100 0000 1001 - 1000 0010 0000 1010 1010 1010 1001 0101 1110 0101 0010 1010 1000 = -1 544.166 661 714 358 269 819 058 477 878 570 556 640 625 May 20 14:33 UTC (GMT)
0 - 011 1111 1100 - 1010 0100 0101 1010 0001 1100 1010 1100 0000 1000 0011 0000 0000 = 0.205 249 999 999 999 488 409 230 252 727 866 172 790 527 343 75 May 20 14:32 UTC (GMT)
1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 = -0.016 738 891 601 562 531 225 022 567 582 527 699 414 640 665 054 321 289 062 5 May 20 14:32 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)