提交 24551f64 编写于 作者: M Michael Ellerman 提交者: Benjamin Herrenschmidt

lmb: Add lmb_free()

We can free memory allocated with lmb_alloc() by removing it from the
list of reserved LMBs. Rework lmb_remove() to allow that possibility
and add lmb_free() which exploits it.

BenH: Removed some useless parenthesis
Signed-off-by: NMichael Ellerman <michael@ellerman.id.au>
Signed-off-by: NBenjamin Herrenschmidt <benh@kernel.crashing.org>
上级 859aefc5
......@@ -42,6 +42,7 @@ extern void __init lmb_init(void);
extern void __init lmb_analyze(void);
extern long lmb_add(u64 base, u64 size);
extern long lmb_remove(u64 base, u64 size);
extern long __init lmb_free(u64 base, u64 size);
extern long __init lmb_reserve(u64 base, u64 size);
extern u64 __init lmb_alloc_nid(u64 size, u64 align, int nid,
u64 (*nid_range)(u64, u64, int *));
......
......@@ -205,9 +205,8 @@ long lmb_add(u64 base, u64 size)
}
long lmb_remove(u64 base, u64 size)
static long __lmb_remove(struct lmb_region *rgn, u64 base, u64 size)
{
struct lmb_region *rgn = &(lmb.memory);
u64 rgnbegin, rgnend;
u64 end = base + size;
int i;
......@@ -254,6 +253,16 @@ long lmb_remove(u64 base, u64 size)
return lmb_add_region(rgn, end, rgnend - end);
}
long lmb_remove(u64 base, u64 size)
{
return __lmb_remove(&lmb.memory, base, size);
}
long __init lmb_free(u64 base, u64 size)
{
return __lmb_remove(&lmb.reserved, base, size);
}
long __init lmb_reserve(u64 base, u64 size)
{
struct lmb_region *_rgn = &lmb.reserved;
......
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