What are the steps to convert
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0100 0100 0000 0010 1100 0110, 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 0100 0100 0000 0010 1100 0110
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 0100 0100 0000 0010 1100 0110(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 + 1 × 2-30 + 0 × 2-31 + 0 × 2-32 + 0 × 2-33 + 1 × 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 + 1 × 2-43 + 0 × 2-44 + 1 × 2-45 + 1 × 2-46 + 0 × 2-47 + 0 × 2-48 + 0 × 2-49 + 1 × 2-50 + 1 × 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 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 931 322 574 615 478 515 625 + 0 + 0 + 0 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0 + 0 + 0 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0 =
0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 =
0.000 000 000 989 687 887 198 442 695 080 302 655 696 868 896 484 375(10)
= 0
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0100 0100 0000 0010 1100 0110, 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.