On Thu, 2015-07-30 at 13:48 +0800, Dongsheng Yang wrote:
> In ubifs, we have to do a budget for inode before marking
> it as dirty. But sometimes, we would call dirty_inode in vfs
> which will not do a budget for inode. In this case, we have
> to do a budget in ubifs_dirty_inode() by ourselvies.
>
> Signed-off-by: Dongsheng Yang <yangds.fnst@xxxxxxxxxxxxxx>
Could you please explain some more the problem you are trying to solve.
Locking looks confusing and broken. It looks like what you are
expressing is that the 'ui_mutex' is optional, and this smells fishy.
> static void ubifs_dirty_inode(struct inode *inode, int flags)
> {
> struct ubifs_inode *ui = ubifs_inode(inode);
> + int locked = mutex_is_locked(&ui->ui_mutex);
Suppose another process has it locked, so 'locked' is set to 1 here.
> + struct ubifs_info *c = inode->i_sb->s_fs_info;
> + int ret = 0;
> +
> + if (!locked)
So we skip this.
> + mutex_lock(&ui->ui_mutex);
And if the other process has released the lock by this time, we do not
mind, right? Therefore, ui_mutex is "optional"?
> - ubifs_assert(mutex_is_locked(&ui->ui_mutex));
> if (!ui->dirty) {
> + if (!locked) {
And similar here, we do not run this code because 'locked' is 1.
> + struct ubifs_budget_req req = { .dirtied_ino
> = 1,
> + .dirtied_ino_d = ALIGN(ui
> ->data_len, 8) };
> + ret = ubifs_budget_space(c, &req);
> + if (ret)
> + goto out;
> + }
But the other process has already released it, and we do not mind?
Please, try to explain what you want to achieve some more. I am not
sure I understand the end goal.
Artem.
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