What are the steps to convert
0 - 1000 0111 - 010 0000 0010 0001 0100 0011, 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:
1000 0111
The last 23 bits contain the mantissa:
010 0000 0010 0001 0100 0011
2. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
1000 0111(2) =
1 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20 =
128 + 0 + 0 + 0 + 0 + 4 + 2 + 1 =
128 + 4 + 2 + 1 =
135(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 = 135 - 127 = 8
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).
010 0000 0010 0001 0100 0011(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 + 1 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 1 × 2-15 + 0 × 2-16 + 1 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 1 × 2-22 + 1 × 2-23 =
0 + 0.25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 976 562 5 + 0 + 0 + 0 + 0 + 0.000 030 517 578 125 + 0 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =
0.25 + 0.000 976 562 5 + 0.000 030 517 578 125 + 0.000 007 629 394 531 25 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =
0.251 015 067 100 524 902 343 75(10)
= 320.259 857 177 734 375
0 - 1000 0111 - 010 0000 0010 0001 0100 0011, a 32 bit single precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (float) = 320.259 857 177 734 375(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.