Binary ↘ Double: The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 000 0000 0000 - 0000 0000 0000 0000 0000 1000 1010 1000 1000 0000 0000 0000 0000 Converted and Written as a Base Ten Decimal System Number (as a Double)
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 1000 1010 1000 1000 0000 0000 0000 0000: 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 0000
The last 52 bits contain the mantissa:
0000 0000 0000 0000 0000 1000 1010 1000 1000 0000 0000 0000 0000
2. Reserved bitpattern.
We notice that all the bits that make up the exponent are on 0 (clear) and at least one bit of the mantissa is set on 1 (set).
This is one of the reserved bitpatterns of the special values of: Denormalized.
Denormalized numbers are too small to be correctly represented so they approximate to zero.
Depending on the sign bit, -0 and +0 are two distinct values though they both compare as equal (0).
3. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
000 0000 0000(2) =
0 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 0 × 20 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =
0(10)
4. 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 = 0 - 1023 = -1023
5. 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 1000 1010 1000 1000 0000 0000 0000 0000(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 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 1 × 2-25 + 0 × 2-26 + 1 × 2-27 + 0 × 2-28 + 1 × 2-29 + 0 × 2-30 + 0 × 2-31 + 0 × 2-32 + 1 × 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 + 0 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 + 0 + 0.000 000 029 802 322 387 695 312 5 + 0 + 0.000 000 007 450 580 596 923 828 125 + 0 + 0.000 000 001 862 645 149 230 957 031 25 + 0 + 0 + 0 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =
0.000 000 476 837 158 203 125 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 007 450 580 596 923 828 125 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 116 415 321 826 934 814 453 125 =
0.000 000 516 069 121 658 802 032 470 703 125(10)
6. 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 + 0.000 000 516 069 121 658 802 032 470 703 125) × 2-1023 =
1.000 000 516 069 121 658 802 032 470 703 125 × 2-1023 =
0
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 1000 1010 1000 1000 0000 0000 0000 0000 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: