I'm working on a task for A-Level Comp Sci and one question involves normalising the number 110.110 in floating point representation with a 6 bit mantissa and a 4 bit exponent. What is the correct way to do this?
I can't move the decimal point to the beginning of the mantissa (0.110110) as that would cause it to exceed 6 bits but I also can't move the decimal point to the first 1 (1.10110) because then it would act as a negative sign bit. Do I just leave it as is with no exponent or is there something I'm missing? Many thanks.
In IEEE-754 normalized numbers mantissa always lies in range 1.0..2.0, in binary 1.0000000 ... 1.1111111. Bit before point is always set, so it is omitted, and we store only part of mantissa after point.
In your case 110.110 = 100 * 1.10110, so mantissa is 10110 (or 101100 if 6 bits needed)
Exponent 100 perhaps is stored as 100 + 111 = 1011 (by analogy with single precision exponent stored as e+127).
Sign bit in IEEE-754 float values is separate leftmost bit. Don't sure about your format
So I suppose the next binary code:
0 1011 101100
s eeee mmmmmm