Binary ↘ Double: The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 000 0000 0001 - 0000 0100 0000 0000 0000 0000 0000 0000 0000 1110 0000 0001 0001 Converted and Written as a Base Ten Decimal System Number (as a Double)
0 - 000 0000 0001 - 0000 0100 0000 0000 0000 0000 0000 0000 0000 1110 0000 0001 0001: 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:
000 0000 0001
The last 52 bits contain the mantissa:
0000 0100 0000 0000 0000 0000 0000 0000 0000 1110 0000 0001 0001
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
000 0000 0001(2) =
0 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =
1 =
1(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 - 1023 = -1022
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 0100 0000 0000 0000 0000 0000 0000 0000 1110 0000 0001 0001(2) =
0 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 1 × 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 + 1 × 2-37 + 1 × 2-38 + 1 × 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 + 1 × 2-48 + 0 × 2-49 + 0 × 2-50 + 0 × 2-51 + 1 × 2-52 =
0 + 0 + 0 + 0 + 0 + 0.015 625 + 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 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0 + 0 + 0 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.015 625 + 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 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.015 625 000 012 736 700 583 104 720 863 047 987 222 671 508 789 062 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 + 0.015 625 000 012 736 700 583 104 720 863 047 987 222 671 508 789 062 5) × 2-1022 =
1.015 625 000 012 736 700 583 104 720 863 047 987 222 671 508 789 062 5 × 2-1022 =
0
0 - 000 0000 0001 - 0000 0100 0000 0000 0000 0000 0000 0000 0000 1110 0000 0001 0001 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: