Question on exam WS17, 3b

Question

Hello,

my question is about exercise 3 b) in the exam of WS 17. I tried to convert the last example -41.1 to ReSyFloat, but I don't understand why the correct solution is NaN.

We have B=2, e=3, m=4 so beta should be 3. I used the algorithm on slides 23-25 in the chapter Floating Point Numbers. In the first step I get E=floor(log_2(41.1))=5, but then we have E+beta = 5+3=8>2^3-1, so we can't represent the exponent using 3 bits and therefore my solution would be overflow and not NaN.

Why is the solution NaN?

Answer 1

Yes, we do not carefully distinguish between overflow and NaN (IEEE 754 does that very carefully). You are right that the problem is rather an overflow, what the teaching tool wants to express is that this number cannot be represented.