compiler: Fix expression switch with non-last default
#4078
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According to https://go.dev/ref/spec#Expression_switches: > There can be at most one default case and it may appear anywhere in the "switch" statement. Previously, `default` statements executed even if they are followed by a matching `case`. Signed-off-by: Leonard Lyubich <[email protected]>
Codecov Report❌ Patch coverage is
Additional details and impacted files@@ Coverage Diff @@
## master #4078 +/- ##
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- Coverage 83.51% 82.68% -0.83%
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Files 351 351
Lines 42404 42422 +18
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- Hits 35412 35075 -337
- Misses 5249 5269 +20
- Partials 1743 2078 +335 ☔ View full report in Codecov by Sentry. 🚀 New features to boost your workflow:
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Turalchik
reviewed
Nov 18, 2025
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| last := len(n.Body.List) - 1 | ||
| for i := range n.Body.List { | ||
| if i < last && len(n.Body.List[i].(*ast.CaseClause).List) == 0 { // early default |
Contributor
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Maybe it's better not to call the len function, but just look at the equality of n.Body.List[i].(*ast.CaseClause).List == nil?
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right. But 1st, @Turalchik do u rly think modifying the original slice is safe enough solution?
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| last := len(n.Body.List) - 1 | ||
| for i := range n.Body.List { |
Contributor
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I think you can use for i := range last {...}, then you won't have to check the condition i < last.
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Problem
i encountered this once while debugging a contract and spent a lot of time searching for the problem
Solution
first, swapping expressions as the most obvious. But it changes the input, and mutation is possible. Cloning is an option, but im gonna think about it some more