Downlevel Omit when used as ExpressionWithTypeArguments #34
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sandersn
requested changes
Aug 3, 2020
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Everything looks correct, but please do refactor to eliminate duplication. I don't think it's worth changing the style of the program yet, so I'd recommend the below changes.
(I'll probably refactor to get closer to a condition -> rewrite style once I get a few more downlevelled features.)
index.js
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| typeArguments[1] | ||
| ]) | ||
| ]); | ||
| } | ||
| } else if (ts.isTypeReferenceNode(n) && ts.isIdentifier(n.typeName) && n.typeName.escapedText === "Omit") { |
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Suggested change
| } else if (ts.isTypeReferenceNode(n) && ts.isIdentifier(n.typeName) && n.typeName.escapedText === "Omit") { | |
| } else if (ts.isTypeReferenceNode(n) && ts.isIdentifier(n.typeName) && n.typeName.escapedText === "Omit" | |
| || ts.isExpressionWithTypeArguments(n) && ts.isIdentifier(n.expression) && n.expression.escapedText === "Omit") { |
index.js
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| typeArguments[1] | ||
| ]) | ||
| ]); | ||
| } | ||
| } else if (ts.isTypeReferenceNode(n) && ts.isIdentifier(n.typeName) && n.typeName.escapedText === "Omit") { | ||
| const symbol = checker.getSymbolAtLocation(n.typeName); |
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Suggested change
| const symbol = checker.getSymbolAtLocation(n.typeName); | |
| const symbol = checker.getSymbolAtLocation(ts.isTypeReferenceNode(n) ? n.typeName : n.expression); |
|
Also, sorry for the long delay -- I only look at downlevel-dts during the RC period for each Typescript release. Let me know if you'd like me to take over the PR if you've timed out. |
|
No worries. I'm afraid I don't have the time to address your feedback this week. If you have the capacity to take this over I would be grateful. |
sandersn
approved these changes
Aug 3, 2020
chrisguttandin
deleted the
downlevel-omit-as-expression-with-type-arguments
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This pull request is intended to fix #33
The code is very similar to the one which down-levels Omit when used in a TypeReferenceNode. Should I refactor it into a function which can be used for both cases?