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SubjectRe: [PATCH 02/21] memblock, numa: Introduce flag into memblock.
Hello,

On Fri, Jul 19, 2013 at 03:59:15PM +0800, Tang Chen wrote:
> +#define MEMBLK_FLAGS_DEFAULT 0x0 /* default flag */

Please don't do this. Just clearing the struct as zero is enough.

> @@ -439,12 +449,14 @@ repeat:
> int __init_memblock memblock_add_node(phys_addr_t base, phys_addr_t size,
> int nid)
> {
> - return memblock_add_region(&memblock.memory, base, size, nid);
> + return memblock_add_region(&memblock.memory, base, size,
> + nid, MEMBLK_FLAGS_DEFAULT);

And just use zero for no flag. Doing something like the above gets
weird with actual flags. e.g. if you add a flag, say, MEMBLK_HOTPLUG,
should it be MEMBLK_FLAGS_DEFAULT | MEMBLK_HOTPLUG or just
MEMBLK_HOTPLUG? If latter, the knowledge that DEFAULT is zero is
implicit, and, if so, why do it at all?

> +static int __init_memblock memblock_reserve_region(phys_addr_t base,
> + phys_addr_t size,
> + int nid,
> + unsigned long flags)
> {
> struct memblock_type *_rgn = &memblock.reserved;
>
> - memblock_dbg("memblock_reserve: [%#016llx-%#016llx] %pF\n",
> + memblock_dbg("memblock_reserve: [%#016llx-%#016llx] with flags %#016lx %pF\n",

Let's please drop "with" and do we really need to print full 16
digits?

Thanks.

--
tejun


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