ast/parser: generate new wildcards for else #5412
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srenatus
commented
Nov 22, 2022
| if v, ok := rule.Head.Args[i].Value.(Var); ok && v.IsWildcard() { | ||
| rule.Head.Args[i].Value = Var(p.genwildcard()) | ||
| } | ||
| } |
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The copy here had been misleading the "wildcard unmangling" logic in the formatter: https://github.com/open-policy-agent/opa/blob/main/format/format.go#L168
It picked up the else's invisible args and counted them as a second use that would necessitate the unmangling. It really doesn't, though, and is best sidestepped by introducing fresh wildcard variables for the else args in the first place.
srenatus
commented
Nov 22, 2022
| @@ -4830,6 +4854,9 @@ func assertParseRule(t *testing.T, msg string, input string, correct *Rule, opts | |||
| if !rule.Head.Ref().Equal(correct.Head.Ref()) { | |||
| t.Errorf("Error on test \"%s\": rule heads not equal: ref = %v (parsed), ref = %v (correct)", msg, rule.Head.Ref(), correct.Head.Ref()) | |||
| } | |||
| if !rule.Head.Equal(correct.Head) { | |||
| t.Errorf("Error on test \"%s\": rule heads not equal: %v (parsed), %v (correct)", msg, rule.Head, correct.Head) | |||
| } | |||
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Just made the comparison a little more helpful. It's not much, but anyways.
The previous behaviour had triggered a check in the formatter for multiple
use of wildcard variables:
f(_) := true { true }
else := false
The formatter found `$1`, the `_` argument of f, again in else, and thus
changed it into `_1`:
f(_1) := true { true }
else := false
There's no extra meaning to the copied wildcards in `else`, they should not
count as second usage.
Signed-off-by: Stephan Renatus <[email protected]>
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srenatus
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Nov 29, 2022
This is a spiritual follow-up to open-policy-agent#5412. A policy like f(_) := 1 { true } { true } { true } would have been pretty-printed as f(_0) := 1 f(_0) := 1 f(_0) := 1 because of the duplicated wildcards. They now get the same treatment as "else" gets: fresh wildcards. Signed-off-by: Stephan Renatus <[email protected]>
srenatus
added a commit
that referenced
this pull request
Nov 29, 2022
This is a spiritual follow-up to #5412. A policy like f(_) := 1 { true } { true } { true } would have been pretty-printed as f(_0) := 1 f(_0) := 1 f(_0) := 1 because of the duplicated wildcards. They now get the same treatment as "else" gets: fresh wildcards. Signed-off-by: Stephan Renatus <[email protected]>
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The previous behaviour had triggered a check in the formatter for multiple use of wildcard variables:
The formatter found
$1, the_argument of f, again in else, and thus changed it into_1:There's no extra meaning to the copied wildcards in
else, they should not count as second usage.Fixes #5347.
elsedefined for a function.