Hi Devin,
had to look this up myself:
you're right, PDU messages air pmt::pair_t; it's a bit hard to read
before, but as soon as you realize car gives you the first element of a
pair, and cdr the second, things become clearer:
d_curr_meta = pmt::car(msg);
d_curr_vect = pmt::cdr(msg);
d_curr_len = pmt::length(d_curr_vect);
So, the second part is the one containing data in a vector, i.e. what
gets copied to the pdu_to_tagged_stream's output as items (see what
happens with d_curr_vect in work()).
Now, what's the structure of d_curr_meta? Looking at l. 96ff:
// Copy tags
if (!pmt::eq(d_curr_meta, pmt::PMT_NIL) ) {
pmt::pmt_t klist(pmt::dict_keys(d_curr_meta));
for (size_t i = 0; i < pmt::length(klist); i++) {
pmt::pmt_t k(pmt::nth(i, klist));
pmt::pmt_t v(pmt::dict_ref(d_curr_meta, k, pmt::PMT_NIL));
add_item_tag(0, nitems_written(0), k, v, alias_pmt());
}
}
we see that d_curr_meta is treated as a dictionary, and the key/pair
values are added as stream tags.
So, you're probably doing it right:
The first part of your pair is the tags you want to have alongside with
your samples, but something is wrong with the way you send samples, so:
what *is* the second part of your pair? Or am I completely
misunderstanding you, and you want to somehow send the contents of your
dictionary as stream items ("samples")?
Cheers,
Marcus
On 21.11.2015 16:45, devin kelly wrote:
> Hello,
>
> I have a block that emits pmt::dict messages. I'd like to connect it
> to a PDU to Tagged Stream block.
>
> My problem is that a dict isn't a pair and I run into this error:
>
> https://github.com/gnuradio/gnuradio/blob/master/gr-blocks/lib/pdu_to_tagged_stream_impl.cc#L63
>
> If I place the dict in a pair, I won't get that error but I do get
> another error telling me my data needs to be a uniform vector.
>
> The dict has multiple data types in it so I'm not sure what the best
> way to proceed is. What's a good way to (a good way to GR I mean)
> structure my messages? Make them pairs where the first part is
> metadata and the second is a uniform vector?
>
> Thanks,
> Devin
>
>
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