0 - 110 1111 1011 - 1101 1010 0100 1000 1100 1110 0100 0110 1000 1100 0000 0000 0110 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

0 - 110 1111 1011 - 1101 1010 0100 1000 1100 1110 0100 0110 1000 1100 0000 0000 0110: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
0 - 110 1111 1011 - 1101 1010 0100 1000 1100 1110 0100 0110 1000 1100 0000 0000 0110, a 64 bit double precision IEEE 754 binary floating point representation standard to decimal?

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

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

The next 11 bits contain the exponent:
110 1111 1011

The last 52 bits contain the mantissa:
1101 1010 0100 1000 1100 1110 0100 0110 1000 1100 0000 0000 0110


2. Convert the exponent from binary (from base 2) to decimal (in base 10).

The exponent is allways a positive integer.

110 1111 1011(2) =

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

1,024 + 512 + 0 + 128 + 64 + 32 + 16 + 8 + 0 + 2 + 1 =

1,024 + 512 + 128 + 64 + 32 + 16 + 8 + 2 + 1 =

1,787(10)

3. Adjust the exponent.

Subtract the excess bits: 2(11 - 1) - 1 = 1023,

that is due to the 11 bit excess/bias notation.

The exponent, adjusted = 1,787 - 1023 = 764


4. Convert the mantissa from binary (from base 2) to decimal (in base 10).

The mantissa represents the fractional part of the number (what comes after the whole part of the number, separated from it by a comma).


1101 1010 0100 1000 1100 1110 0100 0110 1000 1100 0000 0000 0110(2) =

1 × 2-1 + 1 × 2-2 + 0 × 2-3 + 1 × 2-4 + 1 × 2-5 + 0 × 2-6 + 1 × 2-7 + 0 × 2-8 + 0 × 2-9 + 1 × 2-10 + 0 × 2-11 + 0 × 2-12 + 1 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 1 × 2-17 + 1 × 2-18 + 0 × 2-19 + 0 × 2-20 + 1 × 2-21 + 1 × 2-22 + 1 × 2-23 + 0 × 2-24 + 0 × 2-25 + 1 × 2-26 + 0 × 2-27 + 0 × 2-28 + 0 × 2-29 + 1 × 2-30 + 1 × 2-31 + 0 × 2-32 + 1 × 2-33 + 0 × 2-34 + 0 × 2-35 + 0 × 2-36 + 1 × 2-37 + 1 × 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 + 1 × 2-50 + 1 × 2-51 + 0 × 2-52 =

0.5 + 0.25 + 0 + 0.062 5 + 0.031 25 + 0 + 0.007 812 5 + 0 + 0 + 0.000 976 562 5 + 0 + 0 + 0.000 122 070 312 5 + 0 + 0 + 0 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0 + 0 + 0.000 000 014 901 161 193 847 656 25 + 0 + 0 + 0 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0 + 0 + 0 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0 =

0.5 + 0.25 + 0.062 5 + 0.031 25 + 0.007 812 5 + 0.000 976 562 5 + 0.000 122 070 312 5 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 =

0.852 673 427 794 799 376 755 463 526 933 453 977 108 001 708 984 375(10)

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

(-1)Sign × (1 + Mantissa) × 2(Adjusted exponent) =

(-1)0 × (1 + 0.852 673 427 794 799 376 755 463 526 933 453 977 108 001 708 984 375) × 2764 =

1.852 673 427 794 799 376 755 463 526 933 453 977 108 001 708 984 375 × 2764 = ...

= 179 769 313 486 012 321 848 385 327 755 878 509 171 775 370 955 249 762 342 271 040 477 010 192 584 920 747 894 205 664 403 690 014 577 797 118 559 137 269 811 837 453 145 168 980 507 689 930 925 790 096 123 058 873 387 915 795 775 442 510 285 549 816 661 769 046 785 713 785 036 917 210 439 580 582 412 288

0 - 110 1111 1011 - 1101 1010 0100 1000 1100 1110 0100 0110 1000 1100 0000 0000 0110, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 179 769 313 486 012 321 848 385 327 755 878 509 171 775 370 955 249 762 342 271 040 477 010 192 584 920 747 894 205 664 403 690 014 577 797 118 559 137 269 811 837 453 145 168 980 507 689 930 925 790 096 123 058 873 387 915 795 775 442 510 285 549 816 661 769 046 785 713 785 036 917 210 439 580 582 412 288(10)

Spaces were used to group digits: for binary, by 4, for decimal, by 3.

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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)