It's more efficient (especially in terms of memory) to keep the two things contiguously in memory than to have to chase an additional pointer to the array. In C it looks almost the same (arrays and their pointers are both specified by naming the arraym but "sizeof" will act differently for instance.
And yes, the "growable" part of a structure has to be the last thing for it to work.
I learned how to do this in 1981 but scalars only go back to around 1997.
cheers M
On Fri, Feb 21, 2014 at 02:23:43PM -0500, Jonathan Wilkes wrote:
On 02/21/2014 09:00 AM, Ivica Bukvic wrote:
Because this way you can reference data points with sc_vec+n as opposed to dealing with single or double linked lists (since sc_vec can be an array).
If sc_vec is a pointer then you can access data points using the same technique, which is pointer math after all.
For everyone's amusement, here's an exercise in my own rank speculation: something about a t_word array aligning on boundaries in a way that you wouldn't be able to guarantee with a pointer to a t_word. So if you can guarantee there won't be padding you save memory in 1981.
Is it something like that, Miller?
And do scalars actually go back to 1981, or that's just around the time you learned the technique?
Also-- this technique means that for sc_vec[1] its position inside the struct suddenly become relevant. That is, if you put sc_template as the last member field you'd be indexing into the wrong place when you tried to read/write sc_vec data. Is that right?
-Jonathan
On Feb 21, 2014 7:26 AM, "Charles Goyard" <cg@fsck.fr mailto:cg@fsck.fr> wrote:
Hi,
Sorry for this question, but why isn't sc_vec a good old pointer ?
t_gobj sc_gobj; /* header for graphical object */ t_symbol *sc_template; /* template name (LATER replace withpointer) */
t_word sc_vec[1]; /* indeterminate-length array ofwords */
} t_scalar;
How is a static t_word array of size 1 an indeterminate-length
array? Is its placement as the last member of the struct required?
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