In JS Class serialization, read the base construct signature from the static base type#41767
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| /** | ||
| * @param {U} param | ||
| */ | ||
| constructor(param: U); |
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Why is this elided? source has constructor signature for O at https://github.com/microsoft/TypeScript/pull/41767/files#diff-8cef264cf87057b69d50eeda6098081c028ab139d5c75afecc19512c8e92e824R178
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It's identical to the base signature (the base signature is instantiated into the same signature as in the sub class), and thus redundant.
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Interesting.. i thought that would be elided only if O had elided it.
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Nah - the node serializer looks at semantic information usually, not syntactic. Sometimes we reuse input type nodes where possible, but generally not wholesale signature declarations like this. So since the class only has one signature, and it's identical to the signature provided by the base class, we "elide" it so as to not include the base class's signature in all subclasses, since we don't need it.
| const constructors = isNonConstructableClassLikeInJsFile ? | ||
| [factory.createConstructorDeclaration(/*decorators*/ undefined, factory.createModifiersFromModifierFlags(ModifierFlags.Private), [], /*body*/ undefined)] : | ||
| serializeSignatures(SignatureKind.Construct, staticType, baseTypes[0], SyntaxKind.Constructor) as ConstructorDeclaration[]; | ||
| serializeSignatures(SignatureKind.Construct, staticType, staticBaseType, SyntaxKind.Constructor) as ConstructorDeclaration[]; |
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I love the subtlety of this fix and I keep on being amazed that DevTools "fuzzes" TypeScript 😂
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And not the instance base type, where non-static members come from.
Fixes #41397