Hi Spencer,The onset is in bytes, hence the cast to (char *).  I'm just 
wondering why we're measuring things in bytes since an onset must always lie on 
a t_word boundary.
Thanks,Jonathan
 


     On Monday, August 17, 2015 2:29 PM, Jonathan Wilkes <[email protected]> 
wrote:
   

 I guess what I'm saying is that sc_vec is a loaf of bread, and the slices are 
t_word wide.  So why are we doing the bookeeping (i.e., jumping to a particular 
location) in units smaller than one slice?
-Jonathan


     On Monday, August 17, 2015 1:54 PM, Forrest Curo <[email protected]> 
wrote:
   

 *char is guaranteed to be the same size as a 'char', 8 bits -- but if you're 
treating t_word as an array of char's, you can get into t_word and process it 
in 8 bit chunks.

On Mon, Aug 17, 2015 at 9:53 AM, Jonathan Wilkes via Pd-list 
<[email protected]> wrote:

Ah, ok-- I forgot my pointer arithmetic rules.
Once I remembered how it works, I still wasn't understanding how you could be 
doing pointer arithmetic with byte granularity when t_word is wider than 1 
byte.  But then I see from template_find_field you are multiplying the onset by 
sizeof(t_word).
So now, further down the rabbit hole of knowledge, I'm wondering why you 
multiply by sizeof(t_word) at all.  If you didn't, couldn't the (char *) cast 
go away?

-Jonathan
 


     On Monday, August 17, 2015 12:24 PM, Miller Puckette <[email protected]> wrote:
   

 I don't thing the width of (char *) enters into it (it's a pointer, 8 chars in 
64 bit addr space)  - the direct contrast to be made is (char) vs (t_word).

Not sure if that answers the question though...

cheers
M

On Mon, Aug 17, 2015 at 03:52:27PM +0000, Jonathan Wilkes via Pd-list wrote:
> Thanks, I think I'm getting it.
> So is char* guaranteed to be the same width as sizeof(t_word)?  If so, are 
> you just using it as a shorthand?
> 
> Thanks,
> Jonathan
> 
> 
>  
> 
> 
>      On Monday, August 17, 2015 11:31 AM, Miller Puckette <[email protected]> 
>wrote:
>    
> 
>  Here's an example...
> 
> #include <stdio.h>
> 
> float foo[2];
> 
> main()
> {
>     printf("foo = %p\n", foo); 
>     printf("incremented as float: %p\n", foo+1);
>     printf("incremented as (char *): %p\n", ((char *)foo)+1);
> }
> 
> --->
> 
> foo = 0x601038
> incremented as float: 0x60103c
> incremented as (char *): 0x601039
> 
> Adding an integer to a pointer "increments" it - the effect depends on the 
> type
> of pointer.  Another way to think of it is that foo[1], say, is semantically
> identical to *(foo+1).
> 
> cheers
> Miller
> 
> On Mon, Aug 17, 2015 at 03:10:35PM +0000, Jonathan Wilkes via Pd-list wrote:
> > But we're dealing with an array of t_words, so onset*sizeof(t_word) is what 
> > we want anyway, no?
> > -Jonathan 
> > 
> > 
> >      On Monday, August 17, 2015 10:55 AM, Claude Heiland-Allen 
> ><[email protected]> wrote:
> >    
> > 
> >  On 17/08/15 15:36, Jonathan Wilkes via Pd-list wrote:
> > > Hi list,Wondering if someone will give me a free lesson in C programming.
> > > In g_traversal.c, there's some code to retrieve a float from a t_word* 
> > > vec.  It looks like this:
> > > *(t_float *)(((char *)vec) + onset));
> > > Why does vec need to be cast to char*?  t_word has to be as big as the 
> > > largest member of the union, and the largest member has to be the same 
> > > size as char*, right?  (Otherwise we'd have big problems...)
> > 
> > aiui pointer arithmetic is in increments of sizeof(pointee)
> > if onset is measured in bytes (I don't know if it is in this case, but 
> > it looks likely), then you need to have a pointer to bytes for the 
> > addition to be meaningful.  vec is already a pointer, but adding onset 
> > to a t_word* would offset the address by onset*sizeof(t_word) bytes
> > 
> > 
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> > 
> >  
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> 
> 
>  

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