> Is this a known limitation or bug?
it's most likely a feature that tries to protect you from things like:
[-1(
|
[pow 0.5]
|

Ahh yeah makes sense. 

I'm not sure why but I always feel uneasy using [expr], maybe because of libpd :)

[expr] does handle (-1)^0.5 with a NaN output though. Would it be possible to add this to [pow] as well? Something like for negative base values, non-integer exponent values would return NaN?

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?).

Thanks for the reply IOhannes!

Cheers,
Joe   

On 22 April 2013 13:30, IOhannes m zmoelnig <zmoelnig@iem.at> wrote:
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On 2013-04-22 14:19, Joe White wrote:
> Hi,
>
> Just realised putting a negative number into the [pow] object
> outputs '0'?!?
>
> For example if I do:
>
> [-1 ( | [pow 2]
>
> it returns 0, where I would expect it to return 1.
>
> Is this a known limitation or bug?

it's most likely a feature that tries to protect you from things like:

[-1(
|
[pow 0.5]
|

> Are there any work arounds if I want a variable power?

urgh, i had hoped to never have to tell people to use [expr], but
there you go:

[-1\
|
[pack 0 2]
|
[expr pow($f1, $f2)]
|
[1\

vbmdf
IOhannes
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