Re: Question about the "Dynamic reserved memory" patch
> On Fri, 13 Mar 2020 12:06:37 +0900, said:
>
> > In the __reserved_mem_reserve_reg() function, I found something that
> > I couldn't easily understand.
> >
> > To get help, I sent an e-mail to this mailing list.
> >
> > if (first) {
> > fdt_reserved_mem_save_node(node, uname, base, size);
> > first = 0;
> > }
> >
> > I found that fdt_reserved_mem_save_node() is called the regardless of
> > memblock remove/reserve success.
> >
> > I think early_init_dt_reserve_memory_arch() can fail.(ex. for the lack
> > of memblock's region)
> >
> > So I wonder there will be a situation where reserved_mem
> > initialization will be executed without memory reservation.
>
> What you probably missed is that function is wrapped in a #ifdef
> CONFIG_OF_EARLY_FLATTREE - and is called to read in the OF devicetree data and
> save it in a form the kernel can use.
>
> So there usually shouldn't be a problem in reserving memory early in boot,
> unless of course somebody bollixed up a devicetree entry and put in bad values
> for base, size, and nomap.
But, This is when the early memory allocation is not active. Therefore, if
memblock's
region[128] is full, memblock_reserve/remove() can fail.(Of course, there
should be
more than 128 reserved memory.)
>However, fdt_reserved_mem_save_node() needs to happen anyhow, because that's
>not initialiing the memory that wasn't actually reserved, it's recording the
>fact that the devicetree had a reserved memory request in it, and that needs to
>be remembered because there's a second pass over the devicetree data later on
>(or so the comments in drivers/of/of_reserved_mem.c tell me).
Yes, fdt_reserved_mem_save_node() just register the reseved memory in the
reserve_mem.
then, reserved_mem is passed to _reserved_mem_init_node(). But In this
function,
If the node is "cma", it will call rmem_cma_setup() and it will initializing
the cma struct,
So what I was wondering was wouldn't the cma structure be initialized without
memory
being reserved?
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Re: Question about the "Dynamic reserved memory" patch
On Fri, 13 Mar 2020 12:06:37 +0900, said:
> In the __reserved_mem_reserve_reg() function, I found something that
> I couldn't easily understand.
>
> To get help, I sent an e-mail to this mailing list.
> if (first) {
> fdt_reserved_mem_save_node(node, uname, base, size);
> first = 0;
> }
> I found that fdt_reserved_mem_save_node() is called the regardless of
> memblock remove/reserve success.
>
> I think early_init_dt_reserve_memory_arch() can fail.(ex. for the lack
> of memblock's region)
>
> So I wonder there will be a situation where reserved_mem
> initialization will be executed without memory reservation.
What you probably missed is that function is wrapped in a #ifdef
CONFIG_OF_EARLY_FLATTREE - and is called to read in the OF devicetree data and
save it in a form the kernel can use.
So there usually shouldn't be a problem in reserving memory early in boot,
unless of course somebody bollixed up a devicetree entry and put in bad values
for base, size, and nomap. If that happens, the pr_info() call will fire and
hopefully notify somebody there's a problem.
However, fdt_reserved_mem_save_node() needs to happen anyhow, because that's
not initialiing the memory that wasn't actually reserved, it's recording the
fact that the devicetree had a reserved memory request in it, and that needs to
be remembered because there's a second pass over the devicetree data later on
(or so the comments in drivers/of/of_reserved_mem.c tell me).
Having said that, it *may* make sense to elevate the pr_info() call to a
pr_err(), to make it *obvious* that something went pear-shaped in the
devicetree. But that's a decision for the devicetree/OF maintainers.
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Question about the "Dynamic reserved memory" patch
Hi.
Recently, I read about the "Dynamic reserved memory" patch.
In the __reserved_mem_reserve_reg() function, I found something that
I couldn't easily understand.
To get help, I sent an e-mail to this mailing list.
I attached the code below.
static int __init __reserved_mem_reserve_reg(unsigned long node,
const char *uname)
{
int t_len = (dt_root_addr_cells + dt_root_size_cells) * sizeof(__be32);
phys_addr_t base, size;
int len;
const __be32 *prop;
int nomap, first = 1;
prop = of_get_flat_dt_prop(node, "reg", );
if (!prop)
return -ENOENT;
if (len && len % t_len != 0) {
pr_err("Reserved memory: invalid reg property in '%s',skipping
node.\n",
uname);
return -EINVAL;
}
nomap = of_get_flat_dt_prop(node, "no-map", NULL) != NULL;
while (len >= t_len) {
base = dt_mem_next_cell(dt_root_addr_cells, );
size = dt_mem_next_cell(dt_root_size_cells, );
if (size &&
early_init_dt_reserve_memory_arch(base, size, nomap) == 0)
pr_debug("Reserved memory: reserved region for node
'%s': base %pa, size %ld MiB\n",
uname, , (unsigned long)size / SZ_1M);
else
pr_info("Reserved memory: failed to reserve memory for
node '%s': base %pa, size %ld MiB\n",
uname, , (unsigned long)size / SZ_1M);
len -= t_len;
if (first) {
fdt_reserved_mem_save_node(node, uname, base, size);
first = 0;
}
}
return 0;
}
I found that fdt_reserved_mem_save_node() is called the regardless of
memblock remove/reserve success.
I think early_init_dt_reserve_memory_arch() can fail.(ex. for the lack
of memblock's region)
So I wonder there will be a situation where reserved_mem
initialization will be executed without memory reservation.
I would appreciate it if you let me know if I missed anything :)
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