Disclaimer: I don't feel too strongly about this because I realised my original calculation was wrong anyway, however...
If you think about it negative exponents require a completely
different algorithm that is discontinuous with the one for positives. Instead of multiplying you divide n times.
For sure, but we're talking about negative bases, additionally negative exponents are already handled correctly by [pow].
bien sur, it would be easy to add this.
the thing is, do we really want that? having NaN's somewhere in your computation will have all the results become NaN, including any signals. NaN-signals don't sound good.
There are many things I can do in Pd that don't sound good, I've bust my ears and headphones/speakers too many times to know that :) I would argue that yes (I personally) would want that. If the calculation is wrong, then it's my fault. Also, this is in reference to a message object that's being used for a graph calculation. Receiving '0' was confusing.
the output is a number of the value NaN (which gets displayed as NaN, but
this doesn't mean it a symbol ,just like "1e-8" is not a symbol either...usually)
Good to know, thanks!
Out of curiosity, are the workarounds suggested more of a result of the difficulty of extending the Pd core rather than the implications that such a change might have? Obviously [expr] is a good solution but still the fact that [pow] acts differently feels non-intuitive.
Cheers, Joe
On 22 April 2013 21:43, IOhannes zmölnig zmoelnig@iem.at wrote:
On 04/22/2013 04:07 PM, Joe White wrote:
Would it be possible to add this to [pow] as well? Something like for negative base values, non-integer exponent values would return NaN?
bien sur, it would be easy to add this. the thing is, do we really want that? having NaN's somewhere in your computation will have all the results become NaN, including any signals. NaN-signals don't sound good.
Additionally for [pow] to output '0' seems wrong, because that is definitely not the answer. I've never seen NaN output elsewhere so I'm assuming [expr] outputs a symbol and not some Pd defined NaN type (maybe?).
no. the output is a number of the value NaN (which gets displayed as
NaN, but this doesn't mean it a symbol ,just like "1e-8" is not a symbol either...usually)
fmgar IOhannes
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