Float exponent support: "%e" / "%E"

[charlesnicholson]

No description provided.

[charlesnicholson]

Maybe something cheapo like this would work:

Determine a multiplier of 10.0f or 0.1f based on initial condition of 0 <= abs(x) <= 1 or abs(x) > 1. Determine log10(x) by counting how many applications of the multiplier brings x into 1 <= abs(x) < 10. Once x is exponent-normalized just print it like a normal reversed float, but stick the e+02 (or whatever) reversed at the beginning of cbuf.

This means that the concept of precision will have to be pushed into this new etoa_rev function maybe?

[charlesnicholson]

Also make sure that there aren't any floating point numbers where successive multiplications of 0.1 or 10.0 won't eventually normalize it to 1 <= abs(x) < 10. Floats are hard.