1 - 100 0001 1101 - 0010 1001 0000 0100 0100 1110 0001 0000 0000 0000 0000 0010 0110 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
1 - 100 0001 1101 - 0010 1001 0000 0100 0100 1110 0001 0000 0000 0000 0000 0010 0110: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
1 - 100 0001 1101 - 0010 1001 0000 0100 0100 1110 0001 0000 0000 0000 0000 0010 0110, 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.
1
The next 11 bits contain the exponent:
100 0001 1101
The last 52 bits contain the mantissa:
0010 1001 0000 0100 0100 1110 0001 0000 0000 0000 0000 0010 0110
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
100 0001 1101(2) =
1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 =
1,024 + 0 + 0 + 0 + 0 + 0 + 16 + 8 + 4 + 0 + 1 =
1,024 + 16 + 8 + 4 + 1 =
1,053(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 = 1,053 - 1023 = 30
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).
0010 1001 0000 0100 0100 1110 0001 0000 0000 0000 0000 0010 0110(2) =
0 × 2-1 + 0 × 2-2 + 1 × 2-3 + 0 × 2-4 + 1 × 2-5 + 0 × 2-6 + 0 × 2-7 + 1 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 1 × 2-18 + 0 × 2-19 + 0 × 2-20 + 1 × 2-21 + 1 × 2-22 + 1 × 2-23 + 0 × 2-24 + 0 × 2-25 + 0 × 2-26 + 0 × 2-27 + 1 × 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 + 0 × 2-48 + 0 × 2-49 + 1 × 2-50 + 1 × 2-51 + 0 × 2-52 =
0 + 0 + 0.125 + 0 + 0.031 25 + 0 + 0 + 0.003 906 25 + 0 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0.000 003 814 697 265 625 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0 + 0 + 0 + 0 + 0.000 000 003 725 290 298 461 914 062 5 + 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 + 0 + 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 =
0.125 + 0.031 25 + 0.003 906 25 + 0.000 061 035 156 25 + 0.000 003 814 697 265 625 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 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 =
0.160 221 938 043 841 216 625 651 213 689 707 219 600 677 490 234 375(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)1 × (1 + 0.160 221 938 043 841 216 625 651 213 689 707 219 600 677 490 234 375) × 230 =
-1.160 221 938 043 841 216 625 651 213 689 707 219 600 677 490 234 375 × 230 = ...
= -1 245 778 820.000 009 059 906 005 859 375
1 - 100 0001 1101 - 0010 1001 0000 0100 0100 1110 0001 0000 0000 0000 0000 0010 0110, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = -1 245 778 820.000 009 059 906 005 859 375(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.