Binary ↘ Double: The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 011 1111 1111 - 0000 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 Converted and Written as a Base Ten Decimal System Number (as a Double)
0 - 011 1111 1111 - 0000 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1011: 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 1111
The last 52 bits contain the mantissa:
0000 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1011
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
011 1111 1111(2) =
0 × 210 + 1 × 29 + 1 × 28 + 1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20 =
0 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 =
512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 =
1,023(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 = 1,023 - 1023 = 0
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 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1011(2) =
0 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 1 × 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 + 1 × 2-49 + 0 × 2-50 + 1 × 2-51 + 1 × 2-52 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.003 906 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 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.003 906 25 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.003 906 250 000 002 442 490 654 175 344 388 931 989 669 799 804 687 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 + 0.003 906 250 000 002 442 490 654 175 344 388 931 989 669 799 804 687 5) × 20 =
1.003 906 250 000 002 442 490 654 175 344 388 931 989 669 799 804 687 5 × 20 =
1.003 906 250 000 002 442 490 654 175 344 388 931 989 669 799 804 687 5
0 - 011 1111 1111 - 0000 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1011 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) = 1.003 906 250 000 002 442 490 654 175 344 388 931 989 669 799 804 687 5(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: