Fix default property assigned prototype #40836
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The check in getAssignedClassSymbol forgot to allow for default-property assignment declarations, in part because it wasn't using a utility function to do so.
typescript-bot
added
Author: Team
For Milestone Bug
sandersn
commented
Sep 29, 2020
| @@ -24828,7 +24829,7 @@ namespace ts { | |||
| const exprType = checkJsxAttribute(attributeDecl, checkMode); | |||
| objectFlags |= getObjectFlags(exprType) & ObjectFlags.PropagatingFlags; | |||
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| const attributeSymbol = createSymbol(SymbolFlags.Property | SymbolFlags.Transient | member.flags, member.escapedName); | |||
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this is a drive-by fix; createSymbol always adds Transient.
| isVariableDeclaration(decl.parent) && getSymbolOfNode(decl.parent)); | ||
| const prototype = assignmentSymbol && assignmentSymbol.exports && assignmentSymbol.exports.get("prototype" as __String); | ||
| const init = prototype && prototype.valueDeclaration && getAssignedJSPrototype(prototype.valueDeclaration); | ||
| const assignmentSymbol = decl && getSymbolOfExpando(decl, /*allowDeclaration*/ true); |
| return init ? getSymbolOfNode(init) : undefined; | ||
| } | ||
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| function getSymbolOfExpando(node: Node, allowDeclaration?: boolean): Symbol | undefined { |
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changes from the move:
- Return a Symbol instead of a Node, since all callers immediately do that anyway. This is why I moved it into the checker.
- Add a parameter
allowDeclarationthat allowsnodeto be either the expando or its declaration. The parameter disables the checks that assert thatnodeis theexpandoinitialiser. - Add a FunctionDeclaration case when
allowDeclarationis true.
andrewbranch
approved these changes
Sep 29, 2020
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The check in getAssignedClassSymbol forgot to allow for default-property assignment declarations, in part because it wasn't using a utility function to do so.
Fixes #39167