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 8 bits contain the exponent:
1000 0110
The last 23 bits contain the mantissa:
111 0001 0001 0000 0001 1100
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
1000 0110(2) =
1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20 =
128 + 0 + 0 + 0 + 0 + 4 + 2 + 0 =
128 + 4 + 2 =
134(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 = 134 - 127 = 7
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).
111 0001 0001 0000 0001 1100(2) =
1 × 2-1 + 1 × 2-2 + 1 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 1 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 1 × 2-19 + 1 × 2-20 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 =
0.5 + 0.25 + 0.125 + 0 + 0 + 0 + 0.007 812 5 + 0 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0 + 0 =
0.5 + 0.25 + 0.125 + 0.007 812 5 + 0.000 488 281 25 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 =
0.883 304 119 110 107 421 875(10)
= -241.062 927 246 093 75
1 - 1000 0110 - 111 0001 0001 0000 0001 1100 converted from a 32 bit single precision IEEE 754 binary floating point standard representation number to a decimal system number, written in base ten (float) = -241.062 927 246 093 75(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.