Pass the -b flag from the wrapper to the underlying filesystem checker

[mckusick]

Summary

Passing the -b flag from the wrapper to the underlying filesystem checker allows users to specify superblock directly without invoking the underlying checker manually.

See https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=27687 for discussion of this change.

Test Plan

Verify that the -b flag is passed through as requested.

[imp]

Thi

sbin/fsck/fsck.c

173

This syntax looks odd. I wouldn't expect the = in this format. catopt doesn't parse it out and fsck_ffs doesn't take that syntax.

[kib]

sbin/fsck/fsck.c

173

Seconded.

[mckusick]

Respond to Warner's comment.

sbin/fsck/fsck.c

173

You are correct that running

% fsck_ffs -b=1280640

does not work. But running

% fsck -b=1280640

does end up invoking fsck_ffs with the = sign removed, i.e.

% fsck_ffs -b1280640

I do not know where the equal sign is being stripped.

In the interest of clarity I have removed the equal sign.

[kib]

sbin/fsck/fsck.c

173

Apparently it is mangle(), look for the strchr('=') fragment.

So if separated by '=', each part gets its own argument in the argv[] array. But for ' ' (space) it seems to be passed as single item. Does it even work then?

I suspect now that your original patch was correct, sorry for the confusion.