TypeScript学习-Functions
2017-06-30 本文已影响17人
指尖泛出的繁华
Optional and Default Parameters
- 如果默认参数被用作尾参数,那么它起的作用和尾参数是可选的作用是一样的:
function buildName(firstName: string, lastName?: string) {
// ...
}
function buildName(firstName: string, lastName = "Smith") {
// ...
}
(firstName: string, lastName?: string) => string
- 可选参数如果有,那么只能放在尾参数
- 默认参数位置任意,其类型由默认值决定,也就是说如果赋值了,那么值只能是设定的默认值的类型或者
undefined
function buildName(firstName = "sss", lastName: string) {
return firstName + " " + lastName;
}
let result3 = buildName(undefined | string , "ssss");
console.log(result3);
Rest Parameters
-
...
用于无穷个可选参数,用于类型里面定义rest
参数
function buildName(firstName: string, ...restOfName: string[]) {
return firstName + " " + restOfName.join(" ");
}
let buildNameFun: (fname: string, ...rest: string[]) => string = buildName;
this
this是在函数调用时被设定的变量
let deck = {
suits: ["hearts", "spades", "clubs", "diamonds"],
cards: Array(52),
createCardPicker: function() {
return function() {
let pickedCard = Math.floor(Math.random() * 52);
let pickedSuit = Math.floor(pickedCard / 13);
return {suit: this.suits[pickedSuit], card: pickedCard % 13};
}
}
}
let cardPicker = deck.createCardPicker();
let pickedCard = cardPicker();
//出错,this.suits[pickedSuit] this被绑定到了 window | undefined
alert("card: " + pickedCard.card + " of " + pickedCard.suit);
this
和arrow function
- 通过使用
arrow function
绑定函数的上下文 - 可以通过执行加上
--noImplicitThis
来帮助提醒可能存在的绑定上下文错误
let deck = {
suits: ["hearts", "spades", "clubs", "diamonds"],
cards: Array(52),
createCardPicker: function() {
// NOTE: the line below is now an arrow function, allowing us to capture 'this' right here
return () => {
let pickedCard = Math.floor(Math.random() * 52);
let pickedSuit = Math.floor(pickedCard / 13);
return {suit: this.suits[pickedSuit], card: pickedCard % 13};
}
}
}
let cardPicker = deck.createCardPicker();
let pickedCard = cardPicker();
alert("card: " + pickedCard.card + " of " + pickedCard.suit);
this parameter
- 然而,即使
arrow function
绑定了上下文,但是,this.suits[pickedSuit]的类型依然是any
,因为this
来源于对象字面量内部的函数表达式 - 所以TypeScript提供一个
this parameter
来修复这个bug
interface Card {
suit: string;
card: number;
}
interface Deck {
suits: string[];
cards: number[];
createCardPicker(this: Deck): () => Card;
}
let deck: Deck = {
suits: ["hearts", "spades", "clubs", "diamonds"],
cards: Array(52),
// NOTE: The function now explicitly specifies that its callee must be of type Deck
createCardPicker: function(this: Deck) {
return () => {
let pickedCard = Math.floor(Math.random() * 52);
let pickedSuit = Math.floor(pickedCard / 13);
return {suit: this.suits[pickedSuit], card: pickedCard % 13};
}
}
}
let cardPicker = deck.createCardPicker();
let pickedCard = cardPicker();
alert("card: " + pickedCard.card + " of " + pickedCard.suit);
这样就修复了bug,TypeScript知道了this
的类型是Deck
,就没有any
错误了
- this parameter in callbacks
当函数作为参数传入一个一个库中的函数中的时候,这些库会把传入的函数当做普通的函数处理,因此这个函数捕获的this
会变成undefined
,所以在这个函数内涉及到this的会报错。
通过以下几步解决错误,而且可以使用this
- 首先通过
fake parameter this
,确保传入的参数函数是不会捕获this
,第一层检查。
interface UIElement {
addClickListener(onclick: (this: void, e: Event) => void): void;
}
- 然后通过箭头函数来帮忙
class Handler {
info: string;
onClickGood = (e: Event) => { this.info = e.message }
}
以上可以会有缺点。那就是每次传建一个Handler类型的对象都会创建一个新的arrow function
,造成浪费,解决办法是在Handler的原型上创建公用的方法
Overloads
JavaScript函数对于不同的参数可能返回不同类型的对象等等,所以需要如下措施来帮助做type-check
- 通过
overloads
和调用函数相同名字的函数组,来检查每一个可能产生的结果。
let suits = ["hearts", "spades", "clubs", "diamonds"];
function pickCard(x: {suit: string; card: number; }[]): number;
function pickCard(x: number): {suit: string; card: number; };
function pickCard(x): any {
// Check to see if we're working with an object/array
// if so, they gave us the deck and we'll pick the card
if (typeof x == "object") {
let pickedCard = Math.floor(Math.random() * x.length);
return pickedCard;
}
// Otherwise just let them pick the card
else if (typeof x == "number") {
let pickedSuit = Math.floor(x / 13);
return { suit: suits[pickedSuit], card: x % 13 };
}
}
let myDeck = [{ suit: "diamonds", card: 2 }, { suit: "spades", card: 10 }, { suit: "hearts", card: 4 }];
let pickedCard1 = myDeck[pickCard(myDeck)];
alert("card: " + pickedCard1.card + " of " + pickedCard1.suit);
let pickedCard2 = pickCard(15);
alert("card: " + pickedCard2.card + " of " + pickedCard2.suit);