64 bit double precision IEEE 754 binary floating point number 0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0010 0000 0101 1001 0100 0000 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0010 0000 0101 1001 0100 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:
000 0000 0000


The last 52 bits contain the mantissa:
0000 0000 0000 0000 0000 0000 0000 0010 0000 0101 1001 0100 0000

Notice that all the exponent bits are on 0 (clear) and at least one of the mantissa bits is on 1 (set).

This is one of the reserved bitpatterns of the special values of: Denormalized.

Denormalized numbers are too small to be correctly represented so they approximate to zero. Depending on the sign bit, -0 and +0 are two distinct values though they both compare as equal (0).

Proof:

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

000 0000 0000(2) =


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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0(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 = 0 - 1023 = -1023

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):

0000 0000 0000 0000 0000 0000 0000 0010 0000 0101 1001 0100 0000(2) =

0 × 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 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 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 + 0 × 2-24 + 0 × 2-25 + 0 × 2-26 + 0 × 2-27 + 0 × 2-28 + 0 × 2-29 + 0 × 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 + 1 × 2-40 + 1 × 2-41 + 0 × 2-42 + 0 × 2-43 + 1 × 2-44 + 0 × 2-45 + 1 × 2-46 + 0 × 2-47 + 0 × 2-48 + 0 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 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 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 0 + 0 + 0.000 000 000 000 056 843 418 860 808 014 869 689 941 406 25 + 0 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0 + 0 + 0 + 0 + 0 + 0 =


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 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 0.000 000 000 000 056 843 418 860 808 014 869 689 941 406 25 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 =


0.000 000 000 470 734 562 441 066 373 139 619 827 270 507 812 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 + 0.000 000 000 470 734 562 441 066 373 139 619 827 270 507 812 5) × 2-1023 =


1.000 000 000 470 734 562 441 066 373 139 619 827 270 507 812 5 × 2-1023 =


0

0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0010 0000 0101 1001 0100 0000
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =


0(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 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0010 0000 0101 1001 0100 0000 = 0 Sep 23 00:28 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 Sep 23 00:28 UTC (GMT)
1 - 011 1111 1101 - 1100 1100 1100 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -0.449 996 948 242 187 5 Sep 23 00:28 UTC (GMT)
1 - 100 1111 1111 - 1000 1111 0011 0111 0001 1111 1000 1111 0110 0001 1110 0000 0011 = -180 570 220 993 002 741 689 188 481 472 832 370 404 368 915 030 297 950 134 427 552 840 043 433 295 872 Sep 23 00:25 UTC (GMT)
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 0000 = 0 Sep 23 00:25 UTC (GMT)
0 - 100 0111 1110 - 1111 1111 1111 1111 1111 1101 0101 1000 0110 1011 1000 0010 0011 = 340 282 339 999 999 350 154 012 326 571 441 913 856 Sep 23 00:23 UTC (GMT)
0 - 010 0000 1001 - 1110 0110 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 145 064 727 221 844 021 157 009 275 385 754 063 136 972 371 347 977 914 093 581 031 580 862 114 589 613 801 031 008 601 618 288 320 775 381 400 683 326 012 173 268 877 561 452 132 136 746 455 905 935 566 047 1 Sep 23 00:19 UTC (GMT)
0 - 100 0000 0101 - 0011 1001 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 78.437 5 Sep 23 00:17 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 Sep 23 00:15 UTC (GMT)
1 - 000 0000 0001 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 = -0 Sep 23 00:14 UTC (GMT)
1 - 100 0000 0100 - 1011 1011 1011 1011 1010 0100 0101 1010 0000 0101 1000 0000 0000 = -55.466 622 069 638 106 040 656 566 619 873 046 875 Sep 23 00:13 UTC (GMT)
1 - 100 0100 0011 - 0000 1101 0000 0011 0111 1100 1101 0100 0011 0011 0110 1010 1110 = -310 151 591 551 524 798 464 Sep 23 00:09 UTC (GMT)
1 - 101 1110 1110 - 0001 1000 1011 0001 0101 1011 1110 0101 1000 0011 1100 1000 1010 = -112 160 303 098 420 918 950 169 373 941 268 746 235 202 318 350 958 141 201 590 755 275 183 852 985 354 454 512 506 389 707 015 830 140 880 199 611 223 533 365 352 640 767 361 235 614 081 277 231 104 Sep 23 00:08 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)