What are the steps to convert
1 - 1000 1100 - 101 1010 1010 0000 0011 0100, 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:
1000 1100
The last 23 bits contain the mantissa:
101 1010 1010 0000 0011 0100
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
1000 1100(2) =
1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20 =
128 + 0 + 0 + 0 + 8 + 4 + 0 + 0 =
128 + 8 + 4 =
140(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 = 140 - 127 = 13
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).
101 1010 1010 0000 0011 0100(2) =
1 × 2-1 + 0 × 2-2 + 1 × 2-3 + 1 × 2-4 + 0 × 2-5 + 1 × 2-6 + 0 × 2-7 + 1 × 2-8 + 0 × 2-9 + 1 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 1 × 2-18 + 1 × 2-19 + 0 × 2-20 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 =
0.5 + 0 + 0.125 + 0.062 5 + 0 + 0.015 625 + 0 + 0.003 906 25 + 0 + 0.000 976 562 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 =
0.5 + 0.125 + 0.062 5 + 0.015 625 + 0.003 906 25 + 0.000 976 562 5 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 476 837 158 203 125 =
0.708 014 011 383 056 640 625(10)
= -13 992.050 781 25
1 - 1000 1100 - 101 1010 1010 0000 0011 0100, a 32 bit single precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (float) = -13 992.050 781 25(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.