[PD] free C lesson

Forrest Curo treegestalt at gmail.com
Mon Aug 17 19:54:19 CEST 2015


*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 <
pd-list at lists.iem.at> 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 <msp at ucsd.edu> 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 <msp at ucsd.edu>
> 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 <
> claude at mathr.co.uk> 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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