Binary ↘ Double: The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 1 - 000 0000 1111 - 1010 1010 1000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100 Converted and Written as a Base Ten Decimal System Number (as a Double)
1 - 000 0000 1111 - 1010 1010 1000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100: 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.
1
The next 11 bits contain the exponent:
000 0000 1111
The last 52 bits contain the mantissa:
1010 1010 1000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
000 0000 1111(2) =
0 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 8 + 4 + 2 + 1 =
8 + 4 + 2 + 1 =
15(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 = 15 - 1023 = -1008
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).
1010 1010 1000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100(2) =
1 × 2-1 + 0 × 2-2 + 1 × 2-3 + 0 × 2-4 + 1 × 2-5 + 0 × 2-6 + 1 × 2-7 + 0 × 2-8 + 1 × 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 + 1 × 2-47 + 0 × 2-48 + 1 × 2-49 + 1 × 2-50 + 0 × 2-51 + 0 × 2-52 =
0.5 + 0 + 0.125 + 0 + 0.031 25 + 0 + 0.007 812 5 + 0 + 0.001 953 125 + 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 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 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 =
0.5 + 0.125 + 0.031 25 + 0.007 812 5 + 0.001 953 125 + 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 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 =
0.666 015 625 000 009 769 962 616 701 377 555 727 958 679 199 218 75(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)1 × (1 + 0.666 015 625 000 009 769 962 616 701 377 555 727 958 679 199 218 75) × 2-1008 =
-1.666 015 625 000 009 769 962 616 701 377 555 727 958 679 199 218 75 × 2-1008 =
-0
1 - 000 0000 1111 - 1010 1010 1000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1100 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(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: