Marcus & Martin:
I tried the dict_keys() method of checking, but even that can fail.
Here is an example:
[code]
import pmt
d = pmt.make_dict()
d = pmt.dict_add(d, pmt.intern('a'), pmt.intern('a'))
d = pmt.dict_add(d, pmt.intern('b'), pmt.intern('b'))
d = pmt.dict_add(d, pmt.intern('c'), pmt.intern('c'))
a = pmt.cons(d, pmt.make_u8vector(10, 10))
print pmt.dict_keys(a)
[/code]
You end up with: ((c . c))
The dict_keys() method will bomb if there are no elements in the dictionary:
print pmt.dict_keys(pmt.cons(pmt.make_dict(), pmt.make_u8vector(10, 10)))
On Tue, Nov 22, 2016 at 5:55 PM, Dave NotTelling <dmp250...@gmail.com>
wrote:
> Thanks for the explanation!
>
> On Tue, Nov 22, 2016 at 5:29 PM, Marcus Müller <marcus.muel...@ettus.com>
> wrote:
>
>> That's a long story. Essentially, a list is a pair of the first element
>> and a pair of a second element and a pair of the third element and a pair
>> of …
>>
>> Cheers,
>> Marcus
>>
>> On 22.11.2016 23:18, Dave NotTelling wrote:
>>
>> I ask because it feels like a bug. Things like ((a . b), (c . d), (e .
>> f)) are definitely not pairs (assuming a pair is 2 elements) and (in my
>> opinion) should not return true for pmt.is_pair().
>>
>> On Tue, Nov 22, 2016 at 5:12 PM, Dave NotTelling <dmp250...@gmail.com>
>> wrote:
>>
>>> 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/31b28f0cf4694378b2
>>>> 6617616d08b4082668962f/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!
>>>> >
>>>> >
>>>> > _______________________________________________
>>>> > Discuss-gnuradio mailing list
>>>> > Discuss-gnuradio@gnu.org
>>>> > https://lists.gnu.org/mailman/listinfo/discuss-gnuradio
>>>> >
>>>>
>>>>
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>>>
>>>
>>
>>
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