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

[Tejun Heo <tj@kernel.org>]

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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