On Mon, 2007-01-15 at 18:01 -0500, Jimi Xenidis wrote:
> > @@ -530,6 +538,33 @@ static ulong boot_of_alloc(ulong size)
> >
> > pos = pos + i;
> > }
> > +}
> > +
> > +int boot_of_mem_avail(int pos, ulong *startpage, ulong *endpage)
> If you'd like to hide the bitmap, then perhaps the first arg should
> be a start address and return the address of the next "used" page?
That's an idea. However, I was thinking something more along the lines
of an opaque iterator token.
> > +{
> > + ulong freebit;
> > + ulong usedbit;
> > +
> > + /* find first free page. */
> > + freebit = find_next_zero_bit(mem_available_pages,
> > MEM_AVAILABLE_PAGES, pos);
> > + if (freebit >= MEM_AVAILABLE_PAGES) {
> > + /* We know everything after MEM_AVAILABLE_PAGES is still
> > free. */
> > + *startpage = MEM_AVAILABLE_PAGES << PAGE_SHIFT;
> > + *endpage = ~0UL;
> > + return -1;
> > + }
> > + *startpage = freebit << PAGE_SHIFT;
> > +
> > + /* now find first used page after that. */
> > + usedbit = find_next_bit(mem_available_pages,
> > MEM_AVAILABLE_PAGES, freebit);
> > + if (usedbit >= MEM_AVAILABLE_PAGES) {
> > + /* We know everything after MEM_AVAILABLE_PAGES is still
> > free. */
> > + *endpage = ~0UL;
> > + return -1;
> > + }
> > + *endpage = usedbit << PAGE_SHIFT;
>
> I'm not 100% but the code below looks like require that end represent
> a free page so "usedbit - 1"?
It's actually the address of the end of the free memory. For example, if
you have a single page available, let's say at 0x4000, start would be
0x4000 and end would be 0x5000. The length is end - start, or 0x1000.
> > @@ -214,16 +148,22 @@ void memory_init(module_t *mod, int mcou
> >
> > printk("End of RAM: %luMiB (%luKiB)\n", eomem >> 20, eomem >>
> > 10);
> >
> > - /* Architecturally the first 4 pages are exception hendlers, we
> > - * will also be copying down some code there */
> > + /* Architecturally the first 4 pages are exception handlers. */
> > heap_start = 4 << PAGE_SHIFT;
> > - if (oftree < (ulong)_start)
> > - heap_start = ALIGN_UP(oftree_end, PAGE_SIZE);
> > -
>
> all the images below are in the bitmap, so can;t you use
> boot_of_mem_avail() to figure this out?
Yes, and that makes me very happy. :) Revised patch follows.
--
Hollis Blanchard
IBM Linux Technology Center
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