Typescript编译JavaScript后类与extend代码差异问题咨询
这是完全正常的现象,并非系统问题
你看到的代码差异是TypeScript根据编译目标版本生成对应JS代码导致的:
- 当TypeScript的编译目标(
target)设为ES5或更低版本时,由于ES5本身没有原生class语法,TS会把class语法转译为基于构造函数+原型链的ES5兼容代码,也就是你现在看到的function实现形式。 - 如果你希望编译后输出原生JS class语法,只需修改
tsconfig.json中的target配置为ES6(或更高版本,如ES2015、ESNext)。
你的TypeScript代码
class User4 { private city: string = "xyx"; protected number: number = 123123; constructor(public name: string) {} } class additional extends User4 { method1() { console.log(this.number); } } export {};
编译目标为ES5时的输出(你当前看到的结果)
"use strict"; var __extends = (this && this.__extends) || (function () { var extendStatics = function (d, b) { extendStatics = Object.setPrototypeOf || ({ __proto__: [] } instanceof Array && function (d, b) { d.__proto__ = b; }) || function (d, b) { for (var p in b) if (Object.prototype.hasOwnProperty.call(b, p)) d[p] = b[p]; }; return extendStatics(d, b); }; return function (d, b) { if (typeof b !== "function" && b !== null) throw new TypeError("Class extends value " + String(b) + " is not a constructor or null"); extendStatics(d, b); function __() { this.constructor = d; } d.prototype = b === null ? Object.create(b) : (__.prototype = b.prototype, new __()); }; })(); exports.__esModule = true; var User4 = /** @class */ (function () { function User4(name) { this.name = name; this.city = "xyx"; this.number = 123123; } return User4; }()); var additional = /** @class */ (function (_super) { __extends(additional, _super); function additional() { return _super !== null && _super.apply(this, arguments) || this; } additional.prototype.method1 = function () { console.log(this.number); }; return additional; }(User4));
修改编译目标为ES6后的输出示例
如果将tsconfig.json的target设为ES6,编译后的代码会更贴近TS的写法:
"use strict"; exports.__esModule = true; class User4 { constructor(name) { this.name = name; this.city = "xyx"; this.number = 123123; } } class additional extends User4 { method1() { console.log(this.number); } }
内容的提问来源于stack exchange,提问作者BlackWatch021
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