Can anyone offer some explanations? Thanks.
Jiangfan
On Thu, Sep 18, 2008 at 12:16 PM, jiangfan shi <[EMAIL PROTECTED]> wrote:
> Hi, All,
>
> I looked through the gr_chunks_to_symbols_bc code, my understanding is
> that it handles a two-dimension array. Based on the following work()
> function, my understanding is that each input_items[m] is a list of
> chunk (called stream in this function), and the input_items itself is
> a list of stream. Then the function maps each chunk to a complex based
> on the symbol table. Am I right?
>
> =============================
> int
> gr_chunks_to_symbols_bc::work (int noutput_items,
> gr_vector_const_void_star &input_items,
> gr_vector_void_star &output_items)
> {
> assert (noutput_items % d_D == 0);
> assert (input_items.size() == output_items.size());
> int nstreams = input_items.size();
>
> for (int m=0;m<nstreams;m++) {
> const unsigned char *in = (unsigned char *) input_items[m];
> gr_complex *out = (gr_complex *) output_items[m];
>
> // per stream processing
> for (int i = 0; i < noutput_items / d_D; i++){
> assert (((unsigned int)in[i]*d_D+d_D) <= d_symbol_table.size());
> memcpy(out, &d_symbol_table[(unsigned int)in[i]*d_D],
> d_D*sizeof(gr_complex));
> out+=d_D;
> }
> // end of per stream processing
>
> }
> return noutput_items;
> }
>
> ==========================
>
> Another question is that what the gr_chunks_to_symbols_bc constructor
> means? Based on the following code, I do not understand d_D(D) and
> d_symbol_table (symbol_table). I know gr_sync_interpolator() is called
> explicitly in gr_chunks_to_symbols_bc's constructor. are d_D and
> d_symbol_table other classes? or something else?
>
> =========================
> gr_chunks_to_symbols_bc::gr_chunks_to_symbols_bc (const
> std::vector<gr_complex> &symbol_table, const int D)
> : gr_sync_interpolator ("chunks_to_symbols_bc",
> gr_make_io_signature (1, -1, sizeof (unsigned char)),
> gr_make_io_signature (1, -1, sizeof (gr_complex)),
> D),
> d_D (D),
> d_symbol_table (symbol_table)
> {
> }
>
> =========================
>
> Thanks for your help.
>
> Jiangfan
>
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