What are the steps to convert
1 - 000 0111 1000 - 1111 1000 0011 0010 1000 0100 1111 1010 1010 1011 0000 0100 0010, 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:
000 0111 1000
The last 52 bits contain the mantissa:
1111 1000 0011 0010 1000 0100 1111 1010 1010 1011 0000 0100 0010
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
000 0111 1000(2) =
0 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20 =
0 + 0 + 0 + 0 + 64 + 32 + 16 + 8 + 0 + 0 + 0 =
64 + 32 + 16 + 8 =
120(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 = 120 - 1023 = -903
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 1000 0011 0010 1000 0100 1111 1010 1010 1011 0000 0100 0010(2) =
1 × 2-1 + 1 × 2-2 + 1 × 2-3 + 1 × 2-4 + 1 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 1 × 2-12 + 0 × 2-13 + 0 × 2-14 + 1 × 2-15 + 0 × 2-16 + 1 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 1 × 2-22 + 0 × 2-23 + 0 × 2-24 + 1 × 2-25 + 1 × 2-26 + 1 × 2-27 + 1 × 2-28 + 1 × 2-29 + 0 × 2-30 + 1 × 2-31 + 0 × 2-32 + 1 × 2-33 + 0 × 2-34 + 1 × 2-35 + 0 × 2-36 + 1 × 2-37 + 0 × 2-38 + 1 × 2-39 + 1 × 2-40 + 0 × 2-41 + 0 × 2-42 + 0 × 2-43 + 0 × 2-44 + 0 × 2-45 + 1 × 2-46 + 0 × 2-47 + 0 × 2-48 + 0 × 2-49 + 0 × 2-50 + 1 × 2-51 + 0 × 2-52 =
0.5 + 0.25 + 0.125 + 0.062 5 + 0.031 25 + 0 + 0 + 0 + 0 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0 + 0 + 0.000 030 517 578 125 + 0 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 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 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 001 862 645 149 230 957 031 25 + 0 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0 + 0.000 000 000 029 103 830 456 733 703 613 281 25 + 0 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0 + 0 + 0 + 0 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0 =
0.5 + 0.25 + 0.125 + 0.062 5 + 0.031 25 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 030 517 578 125 + 0.000 007 629 394 531 25 + 0.000 000 238 418 579 101 562 5 + 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 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0.000 000 000 029 103 830 456 733 703 613 281 25 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 =
0.969 520 865 629 434 513 977 003 007 312 305 271 625 518 798 828 125(10)
= -0
1 - 000 0111 1000 - 1111 1000 0011 0010 1000 0100 1111 1010 1010 1011 0000 0100 0010, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = -0(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.