What are the steps to convert
0 - 011 1111 1110 - 0111 0101 0010 1000 0010 1000 1110 1001 0000 0010 0110 0100 0000, 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:
011 1111 1110
The last 52 bits contain the mantissa:
0111 0101 0010 1000 0010 1000 1110 1001 0000 0010 0110 0100 0000
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
011 1111 1110(2) =
0 × 210 + 1 × 29 + 1 × 28 + 1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 =
0 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 0 =
512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 =
1,022(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 = 1,022 - 1023 = -1
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).
0111 0101 0010 1000 0010 1000 1110 1001 0000 0010 0110 0100 0000(2) =
0 × 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 + 1 × 2-11 + 0 × 2-12 + 1 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 1 × 2-19 + 0 × 2-20 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 1 × 2-25 + 1 × 2-26 + 1 × 2-27 + 0 × 2-28 + 1 × 2-29 + 0 × 2-30 + 0 × 2-31 + 1 × 2-32 + 0 × 2-33 + 0 × 2-34 + 0 × 2-35 + 0 × 2-36 + 0 × 2-37 + 0 × 2-38 + 1 × 2-39 + 0 × 2-40 + 0 × 2-41 + 1 × 2-42 + 1 × 2-43 + 0 × 2-44 + 0 × 2-45 + 1 × 2-46 + 0 × 2-47 + 0 × 2-48 + 0 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =
0 + 0.25 + 0.125 + 0.062 5 + 0 + 0.015 625 + 0 + 0.003 906 25 + 0 + 0 + 0.000 488 281 25 + 0 + 0.000 122 070 312 5 + 0 + 0 + 0 + 0 + 0 + 0.000 001 907 348 632 812 5 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 + 0 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 007 450 580 596 923 828 125 + 0 + 0.000 000 001 862 645 149 230 957 031 25 + 0 + 0 + 0.000 000 000 232 830 643 653 869 628 906 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0 + 0 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0 + 0 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0 + 0 + 0 + 0 + 0 + 0 =
0.25 + 0.125 + 0.062 5 + 0.015 625 + 0.003 906 25 + 0.000 488 281 25 + 0.000 122 070 312 5 + 0.000 001 907 348 632 812 5 + 0.000 000 476 837 158 203 125 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 007 450 580 596 923 828 125 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 232 830 643 653 869 628 906 25 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 =
0.457 644 040 000 005 247 748 049 441 725 015 640 258 789 062 5(10)
= 0.728 822 020 000 002 623 874 024 720 862 507 820 129 394 531 25
0 - 011 1111 1110 - 0111 0101 0010 1000 0010 1000 1110 1001 0000 0010 0110 0100 0000, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 0.728 822 020 000 002 623 874 024 720 862 507 820 129 394 531 25(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.