64 bit double precision IEEE 754 binary floating point number 0 - 100 0000 0111 - 0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0111 - 0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 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:
100 0000 0111


The last 52 bits contain the mantissa:
0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

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

100 0000 0111(2) =


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


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


1,024 + 4 + 2 + 1 =


1,031(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 = 1,031 - 1023 = 8

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):

0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000(2) =

0 × 2-1 + 1 × 2-2 + 0 × 2-3 + 1 × 2-4 + 1 × 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 + 0 × 2-31 + 0 × 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 + 0 × 2-41 + 0 × 2-42 + 0 × 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 + 0.062 5 + 0.031 25 + 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 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.25 + 0.062 5 + 0.031 25 =


0.343 75(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 + 0.343 75) × 28 =


1.343 75 × 28 =


344

0 - 100 0000 0111 - 0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =


344(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 0000 0111 - 0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 344 Jun 27 12:03 UTC (GMT)
1 - 011 1110 1111 - 0100 1111 1000 1011 0111 0000 0011 1111 0001 1001 1010 1000 0000 = -0.000 020 000 021 546 807 359 579 256 413 439 907 191 786 915 063 858 032 226 562 5 Jun 27 12:02 UTC (GMT)
1 - 101 0000 0000 - 1111 1111 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -462 716 044 100 681 515 305 910 615 874 561 444 273 028 102 784 962 097 704 551 204 844 121 607 897 088 Jun 27 12:01 UTC (GMT)
1 - 100 1010 1110 - 1001 0101 1000 0101 0101 0010 0000 0101 0101 0111 1110 1010 1010 = -75 861 668 866 319 282 910 687 601 504 749 216 720 588 443 188 461 568 Jun 27 12:01 UTC (GMT)
1 - 100 0000 1000 - 1000 0000 1000 1011 1000 0011 1101 0011 0000 0011 1111 1100 0000 = -769.089 960 457 749 839 406 460 523 605 346 679 687 5 Jun 27 12:01 UTC (GMT)
0 - 100 0001 1000 - 1001 0011 0111 0100 1011 1111 1111 1101 0111 1001 0101 1000 0000 = 52 881 791.980 265 617 370 605 468 75 Jun 27 12:01 UTC (GMT)
0 - 100 0001 0001 - 0011 0010 0000 0000 1011 0100 0001 1011 0100 0111 0000 0000 0000 = 313 346.814 164 876 937 866 210 937 5 Jun 27 12:00 UTC (GMT)
0 - 000 0000 0000 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 = 0 Jun 27 12:00 UTC (GMT)
0 - 111 1111 1111 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 = QNaN, Quiet Not a Number Jun 27 12:00 UTC (GMT)
0 - 000 0000 0000 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 = 0 Jun 27 11:59 UTC (GMT)
0 - 100 0011 1110 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 9 223 372 036 854 775 808 Jun 27 11:59 UTC (GMT)
0 - 100 0000 1110 - 0010 0111 1011 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 37 849 Jun 27 11:58 UTC (GMT)
0 - 100 0000 0010 - 0110 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 11.375 Jun 27 11:58 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)