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TS: Fix self-reference check for alias types

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Asger F
2023-04-25 11:27:09 +02:00
parent c3c3faa4b5
commit 799d92b218
1 changed files with 28 additions and 1 deletions
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29
javascript/extractor/lib/typescript/src/type_table.ts
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@@ -450,7 +450,7 @@ export class TypeTable {
this.isInShallowTypeContext ? TypeExtractionState.PendingShallow : TypeExtractionState.PendingFull);
// If the type is the self-type for a named type (not a generic instantiation of it),
// emit the self-type binding for that type.
if (content.startsWith("reference;") && !(isTypeReference(type) && type.target !== type)) {
if (content.startsWith("reference;") && isTypeSelfReference(type)) {
this.selfTypes.symbols.push(this.getSymbolId(type.aliasSymbol || type.symbol));
this.selfTypes.selfTypes.push(id);
}
@@ -1357,3 +1357,30 @@ function isFunctionTypeOrTypeAlias(declaration: ts.Declaration | undefined) {
if (declaration == null) return false;
return declaration.kind === ts.SyntaxKind.FunctionType || declaration.kind === ts.SyntaxKind.TypeAliasDeclaration;
}
/**
* Given a `type` whose type-string is known to be a `reference`, returns true if this is the self-type for the referenced type.
*
* For example, for `type Foo<R> = ...` this returns true if `type` is `Foo<R>`.
*/
function isTypeSelfReference(type: ts.Type) {
if (type.aliasSymbol != null) {
const { aliasTypeArguments } = type;
if (aliasTypeArguments == null) return true;
let declaration = type.aliasSymbol.declarations?.[0];
if (declaration == null || declaration.kind !== ts.SyntaxKind.TypeAliasDeclaration) return false;
let alias = declaration as ts.TypeAliasDeclaration;
for (let i = 0; i < aliasTypeArguments.length; ++i) {
if (aliasTypeArguments[i].symbol?.declarations?.[0] !== alias.typeParameters[i]) {
return false;
}
}
return true;
} else if (isTypeReference(type)) {
return type.target === type;
} else {
// Return true because we know we have mapped this type to kind `reference`, and in the cases
// not covered above (i.e. generic types) it is always a self-reference.
return true;
}
}