bug in (log)

[Daniel Kochmański]

Hey,

sorry for taking my time with the response. While indeed conforming, I've added a separate code path for ratios. I'm conditioning on the integer length (like SBCL does).

static cl_object
ecl_log1_ratio(cl_object x)
{
  cl_object num = x->ratio.num;
  cl_object den = x->ratio.den;
  if (ecl_integer_length(num) != ecl_integer_length(den)) {
    return ecl_minus(ecl_log1(num), ecl_log1(den));
  } else {
    return ecl_log1_fixnum(x);
  }
}

Thanks all for the input. The change is now submitted as a merge request against develop (and will be part of the next release).

Best regards,
Daniel



--
Daniel Kochmański ;; aka jackdaniel | Przemyśl, Poland
TurtleWare - Daniel Kochmański      | www.turtleware.eu

"Be the change that you wish to see in the world." - Mahatma Gandhi



------- Original Message -------
On Sunday, July 10th, 2022 at 5:39 PM, James Cloos <cloos at jhcloos.com> wrote:


> i should have realized this even before noticing ecl's limitation for
> logs of ratios, but at least i finally did....
> 
> since i'monly using log to find the exponent for an arbitrary precision
> replacement for format's ~e, it works to truncate first.
> 
> so i ended up with this:
> 
> (defun ilog (n &optional (base nil base-p))
> "return floor of base b log of n"
> (let ((s (if (< n 1) -1 1))
> (trunc-n (truncate (if (< n 1) (/ n) n))))
> (floor (* s (if base-p
> (log trunc-n base)
> (log trunc-n))))))
> 
> 
> One could use let* and then test s when choosing between n and (/ n), but
> i choose instead to let the compiler optimize the two calls to (< n 1).
> 
> that makes my ~e replacement work exactly as desired on all lisps i've tried.
> 
> (I have a ~// function which chooses output baed on the arg's type, with
> ratios done via that print-e-significant-digits function and everything
> else using write.)
> 
> -JimC
> --
> James Cloos cloos at jhcloos.com OpenPGP: 0x997A9F17ED7DAEA6
> 
>