I was practicing previous midterm questions and I came across a question
regarding floating bit computations (i.e. additions) and it says "Perform
the following floating bit computation, assuming that the operands and
results have 4-bits of precision, but intermediate results use a guard
bit. Show your work, labelling the guard bit."
My solution:
1.101 x 2^3
+1.001 x 2^2
------------
1.2011 x 2^3 => 1.201 x 2^3
What's the guard bit? Thanks.
Well, first you might want to check your binary.
A guard bit is needed to preserve calculation accuracy for a floating
point operation. It was discussed, along with the round bit, in the
lecture.
CS 251 IA
What is the guard bit in this example?
It's on page 20 of the notes. It's not specifically referred to as a
"guard bit" in the notes, but it's just mentioned that you have to keep
that (n+2)th bit there for accuracy in rounding.