0 - 100 0000 0011 - 1010 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
0 - 100 0000 0011 - 1010 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
0 - 100 0000 0011 - 1010 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111, 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 0011
The last 52 bits contain the mantissa:
1010 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
100 0000 0011(2) =
1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20 =
1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 2 + 1 =
1,024 + 2 + 1 =
1,027(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 = 1,027 - 1023 = 4
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).
1010 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111(2) =
1 × 2-1 + 0 × 2-2 + 1 × 2-3 + 0 × 2-4 + 1 × 2-5 + 1 × 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 + 1 × 2-50 + 1 × 2-51 + 1 × 2-52 =
0.5 + 0 + 0.125 + 0 + 0.031 25 + 0.015 625 + 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 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.5 + 0.125 + 0.031 25 + 0.015 625 + 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 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.672 851 562 500 017 541 523 789 077 473 338 693 380 355 834 960 937 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 + 0.672 851 562 500 017 541 523 789 077 473 338 693 380 355 834 960 937 5) × 24 =
1.672 851 562 500 017 541 523 789 077 473 338 693 380 355 834 960 937 5 × 24 = ...
= 26.765 625 000 000 280 664 380 625 239 573 419 094 085 693 359 375
0 - 100 0000 0011 - 1010 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 26.765 625 000 000 280 664 380 625 239 573 419 094 085 693 359 375(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.