Float printf precision
hverbeet at gmail.com
Sun Mar 31 05:45:01 CDT 2013
On 31 March 2013 11:58, Stefan Dösinger <stefandoesinger at gmail.com> wrote:
> I was curious how you ended up with 8 digits. My thinking was that 32
> bit floats have a 23 bit mantissa, and one digit in the printout
> encodes lb(10) bits - so you need ~6,9 digits to encode the mantissa,
> which rounds up to 7. Thinking further, %.6e should do the job as
> well, as there's one additional digit leading the decimal.
> Is my thinking correct, or am I missing something?
It's been a while since I actually did the calculation, but IIRC you
also need to take the implicit extra bit in the significand into
account, so you'd end up with log(2^24) = ~7.2, which would give %.7e.
But I just rounded up to the next "round" number.
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