0 - 100 0000 0101 - 1111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
0 - 100 0000 0101 - 1111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
0 - 100 0000 0101 - 1111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 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 0101
The last 52 bits contain the mantissa:
1111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
100 0000 0101(2) =
1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 1 × 20 =
1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 4 + 0 + 1 =
1,024 + 4 + 1 =
1,029(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 = 1,029 - 1023 = 6
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).
1111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111(2) =
1 × 2-1 + 1 × 2-2 + 1 × 2-3 + 1 × 2-4 + 0 × 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 + 0 × 2-47 + 1 × 2-48 + 1 × 2-49 + 1 × 2-50 + 1 × 2-51 + 1 × 2-52 =
0.5 + 0.25 + 0.125 + 0.062 5 + 0 + 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 + 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 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.5 + 0.25 + 0.125 + 0.062 5 + 0.015 625 + 0.003 906 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 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =
0.957 031 250 000 006 883 382 752 675 970 550 626 516 342 163 085 937 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 + 0.957 031 250 000 006 883 382 752 675 970 550 626 516 342 163 085 937 5) × 26 =
1.957 031 250 000 006 883 382 752 675 970 550 626 516 342 163 085 937 5 × 26 = ...
= 125.250 000 000 000 440 536 496 171 262 115 240 097 045 898 437 5
0 - 100 0000 0101 - 1111 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1111, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 125.250 000 000 000 440 536 496 171 262 115 240 097 045 898 437 5(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.