似乎可以将类嵌套在构造函数中,然后可以从类中的任何位置实例化该类,这是官方的吗?
编辑例如,
class C {
constructor() {
class D {
constructor() { }
}
}
method() {
var a = new D(); // works fine
}
}
//var a = new D(); // fails in outer scope
traceur生成的JS https://google.github.io/traceur-compiler/demo/repl.html
$traceurRuntime.ModuleStore.getAnonymousModule(function() {
"use strict";
var C = function C() {
var D = function D() {};
($traceurRuntime.createClass)(D, {}, {});
};
($traceurRuntime.createClass)(C, {method: function() {
var a = new D();
}}, {});
return {};
});
//# sourceURL=traceured.js
发布于 2018-04-05 03:23:00
像这样的吗?
class A {
constructor () {
this.B = class {
echo () {
console.log('I am B class');
}
}
}
echo () {
this.b = new this.B;
this.b.echo();
}
}
var a = new A;
a.echo();
https://stackoverflow.com/questions/28784375
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