0 - 100 0000 0001 - 1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
0 - 100 0000 0001 - 1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
0 - 100 0000 0001 - 1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011, 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:
100 0000 0001
The last 52 bits contain the mantissa:
1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
100 0000 0001(2) =
1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20 =
1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =
1,024 + 1 =
1,025(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 = 1,025 - 1023 = 2
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).
1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011(2) =
1 × 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 + 1 × 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 + 1 × 2-46 + 0 × 2-47 + 0 × 2-48 + 1 × 2-49 + 0 × 2-50 + 1 × 2-51 + 1 × 2-52 =
0.5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 976 562 5 + 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 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 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 =
0.5 + 0.000 976 562 5 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 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 =
0.500 976 562 500 016 653 345 369 377 348 106 354 475 021 362 304 687 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 + 0.500 976 562 500 016 653 345 369 377 348 106 354 475 021 362 304 687 5) × 22 =
1.500 976 562 500 016 653 345 369 377 348 106 354 475 021 362 304 687 5 × 22 = ...
= 6.003 906 250 000 066 613 381 477 509 392 425 417 900 085 449 218 75
0 - 100 0000 0001 - 1000 0000 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1011, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 6.003 906 250 000 066 613 381 477 509 392 425 417 900 085 449 218 75(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.