linux-ext4 - Re: [PATCH v4 19/19] fs: handle inode->i

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Date:   Mon, 8 Jan 2018 05:30:25 -0800
From:   Matthew Wilcox <willy@...radead.org>
To:     Jeff Layton <jlayton@...nel.org>
Cc:     linux-fsdevel@...r.kernel.org, linux-kernel@...r.kernel.org,
        viro@...iv.linux.org.uk, linux-nfs@...r.kernel.org,
        bfields@...ldses.org, neilb@...e.de, jack@...e.de,
        linux-ext4@...r.kernel.org, tytso@....edu,
        adilger.kernel@...ger.ca, linux-xfs@...r.kernel.org,
        darrick.wong@...cle.com, david@...morbit.com,
        linux-btrfs@...r.kernel.org, clm@...com, jbacik@...com,
        dsterba@...e.com, linux-integrity@...r.kernel.org,
        zohar@...ux.vnet.ibm.com, dmitry.kasatkin@...il.com,
        linux-afs@...ts.infradead.org, dhowells@...hat.com,
        jaltman@...istor.com
Subject: Re: [PATCH v4 19/19] fs: handle inode->i_version more efficiently

On Fri, Dec 22, 2017 at 07:05:56AM -0500, Jeff Layton wrote:
> +	cur = inode_peek_iversion_raw(inode);
> +	for (;;) {
> +		/* If flag is clear then we needn't do anything */
> +		if (!force && !(cur & I_VERSION_QUERIED))
> +			return false;
> +		/* Since lowest bit is flag, add 2 to avoid it */
> +		new = (cur & ~I_VERSION_QUERIED) + I_VERSION_INCREMENT;

Isn't this an extraordinarily complicated way of spelling:

		new = cur + 1;

We know 'cur' has I_VERSION_QUERIED set, so clearing that bit and adding
two is going to be the same as adding 1 ... right?