Binary ↘ Double: The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 011 1111 1110 - 0000 0000 0000 0000 0000 0100 0000 0000 0000 0000 0000 0010 1101 Converted and Written as a Base Ten Decimal System Number (as a Double)
0 - 011 1111 1110 - 0000 0000 0000 0000 0000 0100 0000 0000 0000 0000 0000 0010 1101: 64 bit double precision IEEE 754 binary floating point standard representation number converted to decimal system (base ten)
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:
011 1111 1110
The last 52 bits contain the mantissa:
0000 0000 0000 0000 0000 0100 0000 0000 0000 0000 0000 0010 1101
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
011 1111 1110(2) =
0 × 210 + 1 × 29 + 1 × 28 + 1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 =
0 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 0 =
512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 =
1,022(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 = 1,022 - 1023 = -1
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).
0000 0000 0000 0000 0000 0100 0000 0000 0000 0000 0000 0010 1101(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 + 1 × 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 + 1 × 2-47 + 0 × 2-48 + 1 × 2-49 + 1 × 2-50 + 0 × 2-51 + 1 × 2-52 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.000 000 238 418 579 101 562 5 + 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 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.000 000 238 418 589 093 569 721 626 408 863 812 685 012 817 382 812 5(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 + 0.000 000 238 418 589 093 569 721 626 408 863 812 685 012 817 382 812 5) × 2-1 =
1.000 000 238 418 589 093 569 721 626 408 863 812 685 012 817 382 812 5 × 2-1 =
0.500 000 119 209 294 546 784 860 813 204 431 906 342 506 408 691 406 25
0 - 011 1111 1110 - 0000 0000 0000 0000 0000 0100 0000 0000 0000 0000 0000 0010 1101 converted from a 64 bit double precision IEEE 754 binary floating point standard representation number to a decimal system number, written in base ten (double) = 0.500 000 119 209 294 546 784 860 813 204 431 906 342 506 408 691 406 25(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.
More operations with 64 bit double precision IEEE 754 binary floating point standard representation numbers: