What are the steps to convert
0 - 0011 1101 - 110 0000 0000 0000 0000 0111, a 32 bit single 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 8 bits contain the exponent:
0011 1101
The last 23 bits contain the mantissa:
110 0000 0000 0000 0000 0111
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
0011 1101(2) =
0 × 27 + 0 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 =
0 + 0 + 32 + 16 + 8 + 4 + 0 + 1 =
32 + 16 + 8 + 4 + 1 =
61(10)
3. Adjust the exponent.
Subtract the excess bits: 2(8 - 1) - 1 = 127,
that is due to the 8 bit excess/bias notation.
The exponent, adjusted = 61 - 127 = -66
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).
110 0000 0000 0000 0000 0111(2) =
1 × 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 + 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 + 1 × 2-21 + 1 × 2-22 + 1 × 2-23 =
0.5 + 0.25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 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 =
0.5 + 0.25 + 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 =
0.750 000 834 465 026 855 468 75(10)
= 0.000 000 000 000 000 000 023 716 933 832 230 346 73
0 - 0011 1101 - 110 0000 0000 0000 0000 0111, a 32 bit single precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (float) = 0.000 000 000 000 000 000 023 716 933 832 230 346 73(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.