TypeScript/tests/baselines/reference/apparentTypeSupertype.js
2017-01-03 13:42:12 -08:00

40 lines
1.6 KiB
TypeScript

//// [apparentTypeSupertype.ts]
// subtype checks use the apparent type of the target type
// S is a subtype of a type T, and T is a supertype of S, if one of the following is true, where S' denotes the apparent type (section 3.8.1) of S:
class Base {
x: string;
}
// is String (S) a subtype of U extends String (T)? Would only be true if we used the apparent type of U (T)
class Derived<U extends String> extends Base { // error
x: U;
}
//// [apparentTypeSupertype.js]
// subtype checks use the apparent type of the target type
// S is a subtype of a type T, and T is a supertype of S, if one of the following is true, where S' denotes the apparent type (section 3.8.1) of S:
var __extends = (this && this.__extends) || (function () {
var extendStatics = Object.setPrototypeOf ||
({ __proto__: [] } instanceof Array && function (d, b) { d.__proto__ = b; }) ||
function (d, b) { for (var p in b) if (b.hasOwnProperty(p)) d[p] = b[p]; };
return function (d, b) {
extendStatics(d, b);
function __() { this.constructor = d; }
d.prototype = b === null ? Object.create(b) : (__.prototype = b.prototype, new __());
};
})();
var Base = (function () {
function Base() {
}
return Base;
}());
// is String (S) a subtype of U extends String (T)? Would only be true if we used the apparent type of U (T)
var Derived = (function (_super) {
__extends(Derived, _super);
function Derived() {
return _super !== null && _super.apply(this, arguments) || this;
}
return Derived;
}(Base));