Messages in this thread | | | Subject | Re: memory barrier question | From | James Bottomley <> | Date | Thu, 16 Sep 2010 13:40:43 -0400 |
| |
On Thu, 2010-09-16 at 19:17 +0200, Miklos Szeredi wrote: > On Thu, 16 Sep 2010, James Bottomley wrote: > > On Thu, 2010-09-16 at 18:56 +0200, Miklos Szeredi wrote: > > > On Thu, 16 Sep 2010, Paul E. McKenney wrote: > > > > On Thu, Sep 16, 2010 at 06:06:53PM +0200, Miklos Szeredi wrote: > > > > > On Thu, 16 Sep 2010, Paul E. McKenney wrote: > > > > > > On Thu, Sep 16, 2010 at 03:30:56PM +0100, David Howells wrote: > > > > > > > Miklos Szeredi <miklos@szeredi.hu> wrote: > > > > > > > > > > > > > > > Is the rmb() really needed? > > > > > > > > > > > > > > > > Take this code from fs/namei.c for example: > > > > > > > > > > > > > > > > inode = next.dentry->d_inode; > > > > > > > > if (!inode) > > > > > > > > goto out_dput; > > > > > > > > > > > > > > > > if (inode->i_op->follow_link) { > > > > > > > > > > > > > > > > It happily dereferences dentry->d_inode without a barrier after > > > > > > > > checking it for non-null, while that d_inode might have just been > > > > > > > > initialized on another CPU with a freshly created inode. There's > > > > > > > > absolutely no synchornization with that on this side. > > > > > > > > > > > > > > Perhaps it's not necessary; once set, how likely is i_op to be changed once > > > > > > > I_NEW is cleared? > > > > > > > > > > > > Are the path_get()s protecting this? > > > > > > > > > > No, when creating a file the dentry will go from negative to positive > > > > > independently from lookup. The dentry can get instantiated with an > > > > > inode between the path_get() and dereferencing ->d_inode. > > > > > > > > > > > If there is no protection, then something like rcu_dereference() is > > > > > > needed for the assignment from next.dentry->d_inode. > > > > > > > > > > Do I understand correctly that the problem is that a CPU may have a > > > > > stale cache associated with *inode, one that was loaded before the > > > > > write barrier took effect? > > > > > > > > Yes, especially if the compiler is aggressively optimizing. > > > > > > How do compiler optimizations make a difference? > > > > There are two types of reorderings that cause problems if you expect the > > bus visible ordering to matter. One is CPU issue reordering, where the > > cpu decides to output loads and stores in a different order than the > > input instruction stream actually said. The other is compiler > > re-ordering where the compiler actually reorders the instructions to > > execute in a different order from what you'd expect by simply reading > > the C code. > > > > We have compiler barrier instructions for the latter and barriers which > > issue CPU primitives for the former. > > Right but in the concrete namei example I can't see how a compiler > optimization can make a difference. The order of the loads is quite > clear: > > LOAD inode = next.dentry->inode > if (inode != NULL) > LOAD inode->f_op > > What is there the compiler can optimize?
In this example, it can't the if is conditional on the previous executions, so even a speculating CPU is required to do ordering If the compiler could prove inode wasn't null (which it can't I think in this case) it might then re-order. in your original question, there were ways the compiler could behave strangely (for instance, it would likely start by initialising p to &x rather than NULL).
James
| |