Using grep "devel\s\+3:" to find at least one whitspace is not
portable on all grep versions; not all grep versions understand "\s"
as a "whitespace".
Use a literal TAB followed by SPACE.
The + as a qualifier for "one or more" is not a basic regular
expression; use egrep instead of grep.
Signed-off-by: Torsten Bögershausen <tboegi@web.de>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
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