Re: [Discuss-gnuradio] PMT Oddities

[Martin Braun]

Dave,

pairs pass is_dict(), which is possibly the root cause here. See also:
https://github.com/gnuradio/gnuradio/blob/31b28f0cf4694378b26617616d08b4082668962f/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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