Martin,
Was that done on purpose?
Thank you for the link! I hadn't thought about checking that way.
Thanks!
-Dave
On Tue, Nov 22, 2016 at 5:08 PM, Martin Braun <martin.br...@ettus.com>
wrote:
> Dave,
>
> pairs pass is_dict(), which is possibly the root cause here. See also:
> https://github.com/gnuradio/gnuradio/blob/31b28f0cf4694378b26617616d08b4
> 082668962f/gr-uhd/lib/usrp_block_impl.cc#L487-L494
>
> Cheers,
> M
>
> On 11/22/2016 01:47 PM, Dave NotTelling wrote:
> > I noticed today that the is_dict and is_pair checks are not appearing to
> > work properly. Here is an example that shows the issue:
> >
> > [code]
> >
> > #!/usr/bin/python
> >
> > import pmt
> >
> > def print_pmt(dictVar):
> > print 'isPair:%05s, isDict:%05s, isTuple:%05s => %s' %
> > (pmt.is_pair(dictVar), pmt.is_dict(dictVar), pmt.is_tuple(dictVar),
> dictVar)
> >
> > print 'DICT'
> >
> > d = pmt.make_dict()
> > print_pmt(d)
> >
> > d = pmt.dict_add(d, pmt.intern('a'), pmt.intern('b'))
> > print_pmt(d)
> >
> > d = pmt.dict_add(d, pmt.intern('c'), pmt.intern('d'))
> > print_pmt(d)
> >
> > d = pmt.dict_add(d, pmt.intern('e'), pmt.intern('f'))
> > print_pmt(d)
> >
> > print '\nCONS'
> >
> > p = pmt.cons(pmt.make_dict(), pmt.make_u8vector(0,0))
> > print_pmt(p)
> >
> > [/code]
> >
> > Run that and you'll see what I consider strange behavior. The values of
> > is_pair and is_dict to not match what is expected. Is that by design?
> > If so, why?
> >
> > ((a . b)) is not a pair... It's a single element dictionary
> > ((c . d) (a . b)) i can sorta see this being a pair, but it wasn't
> > created that way
> > ((e . f) (c . d) (a . b)) definitely not a pair as it's 3 elements
> >
> > (() . #[]) don't dictionaries have to be nested?
> >
> >
> > Thanks!
> >
> >
> > _______________________________________________
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> > Discuss-gnuradio@gnu.org
> > https://lists.gnu.org/mailman/listinfo/discuss-gnuradio
> >
>
>
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