Re: [PATCH 4/4] Default to acting like fsck.
From: David Sterba <hidden>
Date: 2014-09-22 13:13:10
On Mon, Sep 22, 2014 at 09:58:34AM +0100, Dimitri John Ledkov wrote:
On 21 September 2014 13:59, Tobias Geerinckx-Rice [off-list ref] wrote:quoted
On 21 September 2014 03:01, Dimitri John Ledkov [off-list ref] wrote:quoted
Inspect arguments, if we are not called as btrfs, then assume we are called to act like fsck.[...]quoted
- if (!strcmp(bname, "btrfsck")) { + if (strcmp(bname, "btrfs") != 0) {That's assuming a lot. Silently (!) breaking people's btrfs-3.15_patched-DontRandomlyPanicV2 is a recipe for needless hair-pulling. Is there a reason for not using something less like strstr(bname, "fsck") that I am missing?Quite. This is verbatim patch as I have currently applied in Debian packaging, and it was a fast fix to prevent breakage we had at one point. Indeed using "strstr(bname, "fsck")" would be better and sufficient to resolve the problem we encountered (specifically fsck.btrfs -> btrfs not acting like btrfs). Also using strstr, would fix btrfsck.my-build to act like fsck tool.
The intention was to provide backward compatibility shortcut for 'btrfsck' -> 'btrfs check' and nothing else. The referenced bug is again for 0.19 but there's an upstream-shipped stub fsck.btrfs (since 3.14) that should avoid any packaging tricks.