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Only emit inferred type-alias if it is fully instantiated

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This commit is contained in:
Kanchalai Tanglertsampan 2016-09-28 14:51:33 -07:00
parent 60ab007d3a
commit 7945eb6f1a
1 changed files with 7 additions and 2 deletions
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9
src/compiler/checker.ts
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@ -2182,9 +2182,14 @@ namespace ts {
// The specified symbol flags need to be reinterpreted as type flags
buildSymbolDisplay(type.symbol, writer, enclosingDeclaration, SymbolFlags.Type, SymbolFormatFlags.None, nextFlags);
}
else if (!(flags & TypeFormatFlags.InTypeAlias) && type.flags & (TypeFlags.Anonymous | TypeFlags.UnionOrIntersection) && type.aliasSymbol &&
else if (!(flags & TypeFormatFlags.InTypeAlias) && ((type.flags & TypeFlags.Anonymous && !(<AnonymousType>type).target) || type.flags & TypeFlags.UnionOrIntersection) && type.aliasSymbol &&
isSymbolAccessible(type.aliasSymbol, enclosingDeclaration, SymbolFlags.Type, /*shouldComputeAliasesToMakeVisible*/ false).accessibility === SymbolAccessibility.Accessible) {
// Only write out inferred type with its corresponding type-alias if type-alias is visible
// We emit inferred type as type-alias at the current localtion if all the following is true
// the input type is has alias symbol that is accessible
// the input type is a union, intersection or anonymous type that is fully instantiated (if not we want to keep dive into)
// e.g.: export type Bar<X, Y> = () => [X, Y];
// export type Foo<Y> = Bar<any, Y>;
// export const y = (x: Foo<string>) => 1 // we want to emit as ...x: () => [any, string])
const typeArguments = type.aliasTypeArguments;
writeSymbolTypeReference(type.aliasSymbol, typeArguments, 0, typeArguments ? typeArguments.length : 0, nextFlags);
}