0 - 100 0001 0100 - 1101 1000 1000 0000 0000 1000 1000 0000 0000 1000 1000 0011 1000 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
0 - 100 0001 0100 - 1101 1000 1000 0000 0000 1000 1000 0000 0000 1000 1000 0011 1000: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
0 - 100 0001 0100 - 1101 1000 1000 0000 0000 1000 1000 0000 0000 1000 1000 0011 1000, 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 0001 0100
The last 52 bits contain the mantissa:
1101 1000 1000 0000 0000 1000 1000 0000 0000 1000 1000 0011 1000
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
100 0001 0100(2) =
1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 1 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 0 × 20 =
1,024 + 0 + 0 + 0 + 0 + 0 + 16 + 0 + 4 + 0 + 0 =
1,024 + 16 + 4 =
1,044(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 = 1,044 - 1023 = 21
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).
1101 1000 1000 0000 0000 1000 1000 0000 0000 1000 1000 0011 1000(2) =
1 × 2-1 + 1 × 2-2 + 0 × 2-3 + 1 × 2-4 + 1 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 1 × 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 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 1 × 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 + 1 × 2-37 + 0 × 2-38 + 0 × 2-39 + 0 × 2-40 + 1 × 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 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =
0.5 + 0.25 + 0 + 0.062 5 + 0.031 25 + 0 + 0 + 0 + 0.001 953 125 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 + 0 + 0.000 000 029 802 322 387 695 312 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0 + 0 + 0 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 25 + 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 + 0 + 0 =
0.5 + 0.25 + 0.062 5 + 0.031 25 + 0.001 953 125 + 0.000 000 476 837 158 203 125 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 000 454 747 350 886 464 118 957 519 531 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 =
0.845 703 631 647 223 730 283 258 191 775 530 576 705 932 617 187 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 + 0.845 703 631 647 223 730 283 258 191 775 530 576 705 932 617 187 5) × 221 =
1.845 703 631 647 223 730 283 258 191 775 530 576 705 932 617 187 5 × 221 = ...
= 3 870 721.062 516 238 540 410 995 483 398 437 5
0 - 100 0001 0100 - 1101 1000 1000 0000 0000 1000 1000 0000 0000 1000 1000 0011 1000, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 3 870 721.062 516 238 540 410 995 483 398 437 5(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.