On Thu, Dec 06, 2018 at 09:56:30AM +0800, [email protected] wrote: > >On Wed, Dec 05, 2018 at 09:15:51AM +0800, Peng Hao wrote: > >> Return 0 when there is enough kvm_mmu_memory_cache object. > >> > >> Signed-off-by: Peng Hao <[email protected]> > >> --- > >> virt/kvm/arm/mmu.c | 2 +- > >> 1 file changed, 1 insertion(+), 1 deletion(-) > >> > >> diff --git a/virt/kvm/arm/mmu.c b/virt/kvm/arm/mmu.c > >> index ed162a6..fcda0ce 100644 > >> --- a/virt/kvm/arm/mmu.c > >> +++ b/virt/kvm/arm/mmu.c > >> @@ -127,7 +127,7 @@ static int mmu_topup_memory_cache(struct > >> kvm_mmu_memory_cache *cache, > >> while (cache->nobjs < max) { > >> page = (void *)__get_free_page(PGALLOC_GFP); > >> if (!page) > >> - return -ENOMEM; > >> + return cache->nobjs >= min ? 0 : -ENOMEM; > > > >This condition will never be true here, as the exact same condition is > >already checked above, and if it had been true, then we wouldn't be here. > > > >What problem are you attempting to solve? > > > if (cache->nobjs >= min) ------here cache->nobjs<min,it > can continue downward > return 0; > while (cache->nobjs < max) { > page = (void *)__get_free_page(PGALLOC_GFP); > if (!page) > return -ENOMEM; -----here it is possible that > (cache->nobjs >= min) and (cache->nobjs<max) > cache->objects[cache->nobjs++] = page; ---here cache->nobjs increasing > } > > I just think the logic of this function is to return 0 as long as > (cache->nobjs >= min). > thanks.
Oh, I see now. This is the case where we can do enough allocating to over the min line, but fail before we get to the max. Thanks, drew
