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Use strict subtype relation in getCommonSupertype (#61903)

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Anders Hejlsberg 2025-06-24 15:38:59 -07:00 committed by GitHub
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commit e6a50a7861
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1 changed files with 11 additions and 1 deletions
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src/compiler/checker.ts
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@ -25340,11 +25340,21 @@ export function createTypeChecker(host: TypeCheckerHost): TypeChecker {
// right is a supertype.
const superTypeOrUnion = literalTypesWithSameBaseType(primaryTypes) ?
getUnionType(primaryTypes) :
reduceLeft(primaryTypes, (s, t) => isTypeSubtypeOf(s, t) ? t : s)!;
getSingleCommonSupertype(primaryTypes);
// Add any nullable types that occurred in the candidates back to the result.
return primaryTypes === types ? superTypeOrUnion : getNullableType(superTypeOrUnion, getCombinedTypeFlags(types) & TypeFlags.Nullable);
}
function getSingleCommonSupertype(types: Type[]) {
// First, find the leftmost type for which no type to the right is a strict supertype, and if that
// type is a strict supertype of all other candidates, return it. Otherwise, return the leftmost type
// for which no type to the right is a (regular) supertype.
const candidate = reduceLeft(types, (s, t) => isTypeStrictSubtypeOf(s, t) ? t : s)!;
return every(types, t => t === candidate || isTypeStrictSubtypeOf(t, candidate)) ?
candidate :
reduceLeft(types, (s, t) => isTypeSubtypeOf(s, t) ? t : s)!;
}
// Return the leftmost type for which no type to the right is a subtype.
function getCommonSubtype(types: Type[]) {
return reduceLeft(types, (s, t) => isTypeSubtypeOf(t, s) ? t : s)!;