[bugfix] callable types suggestion forgot parens on unions and intersections #3342
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src/rules/callableTypesRule.ts
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| let suggestion = `${text.substr(0, colonPos)} =>${text.substr(colonPos + 1)}`; | ||
| if (parent.parent !== undefined && shouldWrapSuggestion(parent.parent)) { |
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.parent can never be undefined here. you can simply assert that it is not undefined with shouldWrapSuggestion(parent.parent!)
| @@ -32,6 +32,16 @@ var fn: {(): void;}; | |||
| ~~~~~~~~~ [type % ('() => void')] | |||
| function f(x: { (): void }): void; | |||
| ~~~~~~~~ [type % ('() => void')] | |||
| function f(x: string | { (): void }): void; | |||
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also add a test where the type is already wrapped in parentheses, e.g. : string | ({ (): void })
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The funny thing is that it still works and doesn't add parens in this case. I guess that if there is already parens, parent.parent doesn't reference the Union, but reference the node with the extra parens.
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That's right, parent.parent is ts.ParenthesizedType in that case. That's why I wanted this test case. To verify that we don't add parens if they are already present.
src/rules/callableTypesRule.ts
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| @@ -46,7 +46,7 @@ export class Rule extends Lint.Rules.AbstractRule { | |||
| function walk(ctx: Lint.WalkContext<void>) { | |||
| return ts.forEachChild(ctx.sourceFile, function cb(node: ts.Node): void { | |||
| if ((isInterfaceDeclaration(node) && noSupertype(node) | |||
| || isTypeLiteralNode(node)) | |||
| || isTypeLiteralNode(node)) | |||
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the alignment here was actually intentional because of the parens. it makes more sense to just remove the line break
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Thanks @cyrilgandon |
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CHANGELOG.md entry:
[bugfix]
callable-typesauto fix produces invalid results