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Improve efficiency and naming

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The unapplied outer parameters check now checks only whether the last
outer parameter is unapplied, and its name is now positive --
are-all-applied vs are-some-unapplied.
This commit is contained in:
Nathan Shively-Sanders 2015-11-10 08:55:24 -08:00
parent eba94b4e85
commit 64e4cc306c
1 changed files with 6 additions and 10 deletions
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16
src/compiler/checker.ts
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@ -2878,7 +2878,7 @@ namespace ts {
let baseType: Type;
const originalBaseType = baseConstructorType && baseConstructorType.symbol ? getDeclaredTypeOfSymbol(baseConstructorType.symbol) : undefined;
if (baseConstructorType.symbol && baseConstructorType.symbol.flags & SymbolFlags.Class &&
!baseTypeHasUnappliedOuterTypeParameters(originalBaseType)) {
areAllOuterTypeParametersApplied(originalBaseType)) {
// When base constructor type is a class with no captured type arguments we know that the constructors all have the same type parameters as the
// class and all return the instance type of the class. There is no need for further checks and we can apply the
// type arguments in the same manner as a type reference to get the same error reporting experience.
@ -2915,20 +2915,16 @@ namespace ts {
}
}
function baseTypeHasUnappliedOuterTypeParameters(type: Type): boolean {
function areAllOuterTypeParametersApplied(type: Type): boolean {
const outerTypeParameters = (<InterfaceType>type).outerTypeParameters;
const typeArguments = (<TypeReference>type).typeArguments;
if (outerTypeParameters) {
// an unapplied type parameter is one
// whose argument symbol is still the same as the parameter symbol
const numParameters = outerTypeParameters.length;
for (let i = 0; i < numParameters; i++) {
if (outerTypeParameters[i].symbol === typeArguments[i].symbol) {
return true;
}
}
const last = outerTypeParameters.length - 1;
const typeArguments = (<TypeReference>type).typeArguments;
return outerTypeParameters[last].symbol !== typeArguments[last].symbol;
}
return false;
return true;
}
function resolveBaseTypesOfInterface(type: InterfaceType): void {