0 - 000 0000 0000 - 0100 0000 0010 0100 0000 0000 0000 0000 0000 0000 0000 0100 1010 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal
0 - 000 0000 0000 - 0100 0000 0010 0100 0000 0000 0000 0000 0000 0000 0000 0100 1010: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal
What are the steps to convert
0 - 000 0000 0000 - 0100 0000 0010 0100 0000 0000 0000 0000 0000 0000 0000 0100 1010, 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:
000 0000 0000
The last 52 bits contain the mantissa:
0100 0000 0010 0100 0000 0000 0000 0000 0000 0000 0000 0100 1010
2. Reserved bitpattern.
Notice that all the bits that make up the exponent are on 0 (clear) and at least one bit of the mantissa is set on 1 (set).
This is one of the reserved bitpatterns of the special values of: Denormalized.
Denormalized numbers are too small to be correctly represented so they approximate to zero.
Depending on the sign bit, -0 and +0 are two distinct values though they both compare as equal (0).
3. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
000 0000 0000(2) =
0 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 0 × 20 =
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =
0(10)
4. 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 = 0 - 1023 = -1023
5. 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).
0100 0000 0010 0100 0000 0000 0000 0000 0000 0000 0000 0100 1010(2) =
0 × 2-1 + 1 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 1 × 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 + 1 × 2-46 + 0 × 2-47 + 0 × 2-48 + 1 × 2-49 + 0 × 2-50 + 1 × 2-51 + 0 × 2-52 =
0 + 0.25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0.000 061 035 156 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 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0 + 0 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 + 0 =
0.25 + 0.000 488 281 25 + 0.000 061 035 156 25 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0.000 000 000 000 000 444 089 209 850 062 616 169 452 667 236 328 125 =
0.250 549 316 406 266 431 300 764 452 316 798 269 748 687 744 140 625(10)
6. 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 + 0.250 549 316 406 266 431 300 764 452 316 798 269 748 687 744 140 625) × 2-1023 =
1.250 549 316 406 266 431 300 764 452 316 798 269 748 687 744 140 625 × 2-1023 = ...
= 0
0 - 000 0000 0000 - 0100 0000 0010 0100 0000 0000 0000 0000 0000 0000 0000 0100 1010, 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.