This is an idea that I have had for a loooong time that I finally got up the nerve to implement. In fact, this idea was one of the first candidates for my float library. It was daunting,but well worth it and easy to generalize to larger precisions

I want to display ##x## as a base-10 string. My starting goal is to get it into a format like ##x=10^{n}y##, where ##n## is an integer. I'm not going to be too picky about ##n##, so long as ##y## is smallish in magnitude. For example, if ##y\in[.1,10]##, then it is a lot easier to convert ##y## to a string and append an e##n## to that string. It would be like a pseudo scientific notation. From there, I can work on the formatting and making it look pretty.

So what I do first, is get an approximate value for ##n##. My floats are already in a format of ##x=2^{e}\cdot m##, so ##n\approx e\cdot log_{10}2##. Since ##e## is an 8-bit signed integer in this case, and we are only looking for an integer result, we can approximate this further by getting ##log_{10}2## to 8 bits of precision. Finally, ##n\approx \lfloor \frac{77e}{256}\rfloor##. In assembly, this translates to "Multiply by 77 and take the overflow."

Now that I have ##n##, I want to compute ##y## This is done as ##y=\frac{x}{10^{n}}##. Luckily, this sort of exponentiation is easy. For example, ##10^{13}=10^{1+4+8}=10^{1}\cdot 10^{4}\cdot 10^{8}##. In this case, ##0\le n\le 38##, so if we precompute a table for ##10^{1},10^{2},10^{4},10^{8},10^{16},10^{32}##, then we will only ever need at most 5 multiplications to compute ##10^{n}##. With this, we compute ##y##, which should be on ##y\in [.5,20]##. From there, we proceed to convert y to a string and perform any fancy formatting our hearts desire.


Applying this algorithm to the 80-bit floats will make it a lot faster to convert. It won't be as fast as converting a BCD float like TI's, but it will be more on the order of 40,000cc instead of 400,000cc