Em quarta-feira, 9 de fevereiro de 2011, às 04:29:28, Boyd Stephen Smith Jr. escreveu: > A-*---*-B > \ / > *-*-* > A = Good > B = Bad > > Yes, no middle point, and if the commits were linear, then there would be > exactly one to choose. However, there are other cases where a non-linear > history allows git bisect to test fewer commits, instead of > more. (Basically, almost any time a merge commit and the merge-base of > all it's parents are either good or bad.)
Right. If A is good and B is good, then all intervening commits should be good
too. Similarly if both are bad.
But the problem is that you had to test both A and B first to find that out.
Imagine this:
good--*--A--*--*--*--*--Z--X--*--*--*--*--*---B--bad
\ /
*-----*--------Y-----*-----*-----*
Suppose X marks the spot and Y and Z are innocent. In order to bisect this and
find X, knowing the two boundaries, you need to check that B is bad and A is
good.
You can't choose a middle-point in either side of the branch: if you choose
any commit in the AYB branch, all will test "good". If you chose Z, it would
also test "good".
Instead, you have to test both of B^1 and B^2. You have to identify first which
side of the branch has the fault -- and note that both could be good, in which
case the merge is the fault. (I guess it actually doesn't test all parents,
just N-1 and then moves on to normal linear bisecting -- the merge commit
being the faulty one would be a worst-case scenario, but also an unlikely one)
The point is that whenever you have a fault between a branch and a merge, you
need to stop bisecting and instead exclude branches. That is not bisecting and
could, under circumstances, require more tests than proper bisecting.
--
Thiago Macieira - thiago (AT) macieira.info - thiago (AT) kde.org
Senior Product Manager - Nokia, Qt Development Frameworks
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