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", &len);
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, &prop);
size = dt_mem_next_cell(dt_root_size_cells, &prop);
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, &base, (unsigned long)size / SZ_1M);
else
pr_info("Reserved memory: failed to reserve memory for node '%s': base %pa, size %ld MiB\n",
uname, &base, (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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