2014-07-13 01:04:16 +02:00
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//// [apparentTypeSubtyping.ts]
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// subtype checks use the apparent type of the target type
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// 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:
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class Base<U extends String> {
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x: U;
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}
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// is String (S) a subtype of U extends String (T)? Would only be true if we used the apparent type of U (T)
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class Derived<U> extends Base<string> { // error
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x: String;
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}
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class Base2 {
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x: String;
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static s: String;
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}
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// is U extends String (S) a subtype of String (T)? Apparent type of U is String so it succeeds
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class Derived2<U extends String> extends Base2 { // error because of the prototype's not matching, not because of the instance side
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x: U;
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}
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//// [apparentTypeSubtyping.js]
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2014-08-19 19:43:13 +02:00
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// subtype checks use the apparent type of the target type
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// 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:
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2014-07-13 01:04:16 +02:00
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var __extends = this.__extends || function (d, b) {
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for (var p in b) if (b.hasOwnProperty(p)) d[p] = b[p];
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function __() { this.constructor = d; }
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__.prototype = b.prototype;
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d.prototype = new __();
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};
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var Base = (function () {
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function Base() {
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}
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return Base;
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})();
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2014-08-14 15:42:18 +02:00
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// is String (S) a subtype of U extends String (T)? Would only be true if we used the apparent type of U (T)
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2014-07-13 01:04:16 +02:00
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var Derived = (function (_super) {
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__extends(Derived, _super);
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function Derived() {
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_super.apply(this, arguments);
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}
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return Derived;
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})(Base);
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var Base2 = (function () {
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function Base2() {
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}
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return Base2;
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})();
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2014-08-14 15:42:18 +02:00
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// is U extends String (S) a subtype of String (T)? Apparent type of U is String so it succeeds
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2014-07-13 01:04:16 +02:00
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var Derived2 = (function (_super) {
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__extends(Derived2, _super);
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function Derived2() {
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_super.apply(this, arguments);
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}
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return Derived2;
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})(Base2);
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