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Message-ID: <4CABDF3A.2020501@zytor.com>
Date: Tue, 05 Oct 2010 19:30:18 -0700
From: "H. Peter Anvin" <hpa@...or.com>
To: Yinghai Lu <yinghai@...nel.org>
CC: Benjamin Herrenschmidt <benh@...nel.crashing.org>,
Ingo Molnar <mingo@...e.hu>,
Thomas Gleixner <tglx@...utronix.de>,
LKML <linux-kernel@...r.kernel.org>
Subject: Re: Section mismatches in memblock
On 10/05/2010 04:30 PM, Yinghai Lu wrote:
>
> looks like gcc problem.
What makes you say that?
> | static int memblock_search(struct memblock_type *type, phys_addr_t addr)
> | {
> | .....
> | }
> |
> |int __init memblock_is_reserved(phys_addr_t addr)
> |{
> | return memblock_search(&memblock.reserved, addr) != -1;
> |}
> |
> |int __init_memblock memblock_is_memory(phys_addr_t addr)
> |{
> | return memblock_search(&memblock.memory, addr) != -1;
> |}
>
> Because We already have __init for those two functions.
>
> Just make memblock_search to have __init_memblock atribute.
>
Wrong functions! Furhtermore, at least the tip tree definitely does not
have __init_memblock here:
static int memblock_search(struct memblock_type *type, phys_addr_t addr)
{
unsigned int left = 0, right = type->cnt;
do {
unsigned int mid = (right + left) / 2;
if (addr < type->regions[mid].base)
right = mid;
else if (addr >= (type->regions[mid].base +
type->regions[mid].size))
left = mid + 1;
else
return mid;
} while (left < right);
return -1;
}
int __init memblock_is_reserved(phys_addr_t addr)
{
return memblock_search(&memblock.reserved, addr) != -1;
}
int memblock_is_memory(phys_addr_t addr)
{
return memblock_search(&memblock.memory, addr) != -1;
}
int memblock_is_region_memory(phys_addr_t base, phys_addr_t size)
{
int idx = memblock_search(&memblock.reserved, base);
if (idx == -1)
return 0;
return memblock.reserved.regions[idx].base <= base &&
(memblock.reserved.regions[idx].base +
memblock.reserved.regions[idx].size) >= (base + size);
}
... so I don't know why you're saying that you already have them.
-hpa
--
H. Peter Anvin, Intel Open Source Technology Center
I work for Intel. I don't speak on their behalf.
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