What are the steps to convert
1 - 0010 1101 - 110 1110 1110 1000 0100 0001, 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.
1
The next 8 bits contain the exponent:
0010 1101
The last 23 bits contain the mantissa:
110 1110 1110 1000 0100 0001
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
0010 1101(2) =
0 × 27 + 0 × 26 + 1 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 =
0 + 0 + 32 + 0 + 8 + 4 + 0 + 1 =
32 + 8 + 4 + 1 =
45(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 = 45 - 127 = -82
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 1110 1110 1000 0100 0001(2) =
1 × 2-1 + 1 × 2-2 + 0 × 2-3 + 1 × 2-4 + 1 × 2-5 + 1 × 2-6 + 0 × 2-7 + 1 × 2-8 + 1 × 2-9 + 1 × 2-10 + 0 × 2-11 + 1 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 1 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 1 × 2-23 =
0.5 + 0.25 + 0 + 0.062 5 + 0.031 25 + 0.015 625 + 0 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0 + 0.000 244 140 625 + 0 + 0 + 0 + 0 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0 + 0 + 0.000 000 119 209 289 550 781 25 =
0.5 + 0.25 + 0.062 5 + 0.031 25 + 0.015 625 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 244 140 625 + 0.000 007 629 394 531 25 + 0.000 000 119 209 289 550 781 25 =
0.866 462 826 728 820 800 781 25(10)
= -0
1 - 0010 1101 - 110 1110 1110 1000 0100 0001, a 32 bit single precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (float) = -0(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.