How to manually (bitwise) perform (float)x?

Question

Now, here is the function header of the function I\'m supposed to implement:

/*
 * float_from_int - Return bit-level equivalent of expression (float) x
 *            


        

Answer 1

The basic formulation of the algorithm is to determine the sign, exponent and mantissa bits, then pack the result into an integer. Breaking it down this way makes it easy to clearly separate the tasks in code and makes solving the problem (and testing your algorithm) much easier.

The sign bit is the easiest, and getting rid of it makes finding the exponent easier. You can distinguish four cases: 0, 0x80000000, [-0x7ffffff, -1], and [1, 0x7fffffff]. The first two are special cases, and you can trivially get the sign bit in the last two cases (and the absolute value of the input). If you're going to cast to unsigned, you can get away with not special-casing 0x80000000 as I mentioned in a comment.

Next up, find the exponent -- there's an easy (and costly) looping way, and a trickier but faster way to do this. My absolute favourite page for this is Sean Anderson's bit hacks page. One of the algorithms shows a very quick loop-less way to find the log₂ of an integer in only seven operations.

Once you know the exponent, then finding the mantissa is easy. You just drop the leading one bit, then shift the result either left or right depending on the exponent's value.

If you use the fast log₂ algorithm, you can probably end up with an algorithm which uses no more than 20 operations.