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Unwrap parens when checking for JSDocFunctionType in conditional expression (#46962)

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Jake Bailey
2021-12-06 16:44:37 -08:00
committed by GitHub
parent 4761ba6285
commit 4013271cd8
5 changed files with 25 additions and 2 deletions
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10
src/compiler/parser.ts
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@@ -4539,10 +4539,16 @@ namespace ts {
// - "(x,y)" is a comma expression parsed as a signature with two parameters.
// - "a ? (b): c" will have "(b):" parsed as a signature with a return type annotation.
// - "a ? (b): function() {}" will too, since function() is a valid JSDoc function type.
// - "a ? (b): (function() {})" as well, but inside of a parenthesized type.
// - "a ? (b): (function() {})" as well, but inside of a parenthesized type with an arbitrary amount of nesting.
//
// So we need just a bit of lookahead to ensure that it can only be a signature.
const hasJSDocFunctionType = type && (isJSDocFunctionType(type) || isParenthesizedTypeNode(type) && isJSDocFunctionType(type.type));
let unwrappedType = type;
while (unwrappedType?.kind === SyntaxKind.ParenthesizedType) {
unwrappedType = (unwrappedType as ParenthesizedTypeNode).type; // Skip parens if need be
}
const hasJSDocFunctionType = unwrappedType && isJSDocFunctionType(unwrappedType);
if (!allowAmbiguity && token() !== SyntaxKind.EqualsGreaterThanToken && (hasJSDocFunctionType || token() !== SyntaxKind.OpenBraceToken)) {
// Returning undefined here will cause our caller to rewind to where we started from.
return undefined;