What are the steps to convert
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1100 1001 0001, a 64 bit double precision IEEE 754 binary floating point representation standard to decimal?
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 0000 0000 0000 0000 0100 1100 1001 0001
2. Reserved bitpattern.
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 0000 0000 0000 0000 0100 1100 1001 0001(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 + 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 + 1 × 2-38 + 0 × 2-39 + 0 × 2-40 + 1 × 2-41 + 1 × 2-42 + 0 × 2-43 + 0 × 2-44 + 1 × 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 + 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 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0 + 0 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 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 =
0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 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 =
0.000 000 000 004 352 296 301 135 538 669 768 720 865 249 633 789 062 5(10)
= 0
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1100 1001 0001, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 0(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.