0 - 100 0000 0100 - 1010 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
0 - 100 0000 0100 - 1010 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
0 - 100 0000 0100 - 1010 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101, 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 0100
The last 52 bits contain the mantissa:
1010 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
100 0000 0100(2) =
1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 0 × 20 =
1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 4 + 0 + 0 =
1,024 + 4 =
1,028(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 = 1,028 - 1023 = 5
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 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101(2) =
1 × 2-1 + 0 × 2-2 + 1 × 2-3 + 0 × 2-4 + 1 × 2-5 + 1 × 2-6 + 0 × 2-7 + 1 × 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 + 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 + 1 × 2-47 + 1 × 2-48 + 1 × 2-49 + 1 × 2-50 + 0 × 2-51 + 1 × 2-52 =
0.5 + 0 + 0.125 + 0 + 0.031 25 + 0.015 625 + 0 + 0.003 906 25 + 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 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 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 + 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 + 0.003 906 25 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 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 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.675 781 250 000 013 544 720 900 426 909 793 168 306 350 708 007 812 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 + 0.675 781 250 000 013 544 720 900 426 909 793 168 306 350 708 007 812 5) × 25 =
1.675 781 250 000 013 544 720 900 426 909 793 168 306 350 708 007 812 5 × 25 = ...
= 53.625 000 000 000 433 431 068 813 661 113 381 385 803 222 656 25
0 - 100 0000 0100 - 1010 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1101, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 53.625 000 000 000 433 431 068 813 661 113 381 385 803 222 656 25(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.