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t9107: use shell parameter expansion to avoid breaking &&-chain
This test intentionally breaks the &&-chain when using `expr` to parse
"[<path>]:<ref>" since the pattern matching operation will return 1
(failure) when <path> is empty even though an empty <path> is legitimate
in this test and should not cause the test to fail. However, it is
possible to parse the input without breaking the &&-chain by using shell
parameter expansion (i.e. `${i%%...}`). Other ways to avoid the problem
would be `{ expr $i : ... ||:; }` or test_might_fail(), however,
parameter expansion seems simplest.
IMPLEMENTATION NOTE
The rewritten `if` expression:
if test "$ref" = "${ref#refs/remotes/}"`; then continue; fi
is perhaps a bit subtle. At first glance, it looks like it will
`continue` the loop if $ref starts with "refs/remotes/", but in fact
it's the opposite: the loop will `continue` if $ref does not start with
"refs/remotes/".
In the original, `expr` would only match if the ref started with
"refs/remotes/", and $ref would end up empty if it didn't, so `test -z`
would `continue` the loop if the ref did not start with "refs/remotes/".
With parameter expansion, ${ref#refs/remotes/} attempts to strip
"refs/remotes/" from $ref. If it fails, meaning that $ref does not start
with "refs/remotes/", then the expansion will just be $ref unchanged,
and it will `continue` the loop. On the other hand, if stripping
succeeds, meaning that $ref begins with "refs/remotes/", then the
expansion will be the value of $ref with "refs/remotes/" removed, hence
`continue` will not be taken.
Signed-off-by: Eric Sunshine <[email protected]>
Reviewed-by: Elijah Newren <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>
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