Re: [PATCH 06/18] t5613: clarify "too deep" recursion tests

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On Tue, Oct 4, 2016 at 6:48 AM, Jeff King <peff@xxxxxxxx> wrote:
> On Mon, Oct 03, 2016 at 10:57:48PM -0700, Jacob Keller wrote:
>
>> > diff --git a/t/t5613-info-alternate.sh b/t/t5613-info-alternate.sh
>> > index 7bc1c3c..b393613 100755
>> > --- a/t/t5613-info-alternate.sh
>> > +++ b/t/t5613-info-alternate.sh
>> > @@ -39,6 +39,18 @@ test_expect_success 'preparing third repository' '
>> >         )
>> >  '
>> >
>> > +# Note: These tests depend on the hard-coded value of 5 as "too deep". We start
>> > +# the depth at 0 and count links, not repositories, so in a chain like:
>> > +#
>> > +#   A -> B -> C -> D -> E -> F -> G -> H
>> > +#      0    1    2    3    4    5    6
>> > +#
>>
>> Ok so we count links, but wouldn't we have 5 links when we hit F, and
>> not G? Or am I missing something here?
>
> This is what I was trying to get at with the "start the depth at 0". We
> disallow a depth greater than 5, but because we start at 0-counting,
> it's really six links. I guess saying "5 as too deep" is really the
> misleading part. It should be "5 as the maximum depth".
>
> -Peff

Right, but if A is 0, then:

B = 1
C = 2
D = 3
E = 4
F = 5
G = 6  (UhOh??)
H = 7

So do you mean that *B* = 0, and C = 1??? That is not clear from this commment.

So either way it still feels like "6" links is what is allowed? Or the
first link has to not count? That's really confusing.

Basically I G is the 7th letter, not the 6th, so even if we're
subtractnig 1 it's still 6 which is 1 too deep? That means we not only
discard 0 (the first repository) but we discount the 2nd one as well?

Thanks,
Jake



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