[RFC PATCH 12/29] memblock: replace alloc_bootmem_low with memblock_alloc_low

Mike Rapoport
2018-09-05T15:59:27Z

The alloc_bootmem_low(size) allocates low memory with default alignement
and can be replcaed by memblock_alloc_low(size, 0)

Signed-off-by: Mike Rapoport <rppt@linux.vnet.ibm.com>
---
arch/arm64/kernel/setup.c | 2 +-
arch/unicore32/kernel/setup.c | 2 +-
2 files changed, 2 insertions(+), 2 deletions(-)

diff --git a/arch/arm64/kernel/setup.c b/arch/arm64/kernel/setup.c
index 5b4fac4..cf7a7b7 100644
--- a/arch/arm64/kernel/setup.c
+++ b/arch/arm64/kernel/setup.c
@@ -213,7 +213,7 @@ static void __init request_standard_resources(void)
kernel_data.end = __pa_symbol(_end - 1);

for_each_memblock(memory, region) {
- res = alloc_bootmem_low(sizeof(*res));
+ res = memblock_alloc_low(sizeof(*res), 0);
if (memblock_is_nomap(region)) {
res->name = "reserved";
res->flags = IORESOURCE_MEM;
diff --git a/arch/unicore32/kernel/setup.c b/arch/unicore32/kernel/setup.c
index c2bffa5..9f163f9 100644
--- a/arch/unicore32/kernel/setup.c
+++ b/arch/unicore32/kernel/setup.c
@@ -207,7 +207,7 @@ request_standard_resources(struct meminfo *mi)
if (mi->bank[i].size = 0)
continue;

- res = alloc_bootmem_low(sizeof(*res));
+ res = memblock_alloc_low(sizeof(*res), 0);
res->name = "System RAM";
res->start = mi->bank[i].start;
res->end = mi->bank[i].start + mi->bank[i].size - 1;
--
2.7.4