TypeScript學習-Functions

Optional and Default Parameters

  • 如果默認參數被用作尾參數,那么它起的作用和尾參數是可選的作用是一樣的:
function buildName(firstName: string, lastName?: string) {
    // ...
}
function buildName(firstName: string, lastName = "Smith") {
    // ...
}
(firstName: string, lastName?: string) => string
  • 可選參數如果有,那么只能放在尾參數
  • 默認參數位置任意,其類型由默認值決定,也就是說如果賦值了,那么值只能是設定的默認值的類型或者undefined
function buildName(firstName = "sss", lastName: string) {
    return firstName + " " + lastName;
}
let result3 = buildName(undefined | string , "ssss");
console.log(result3);

Rest Parameters

  • ...用于無窮個可選參數,用于類型里面定義rest參數
function buildName(firstName: string, ...restOfName: string[]) {
    return firstName + " " + restOfName.join(" ");
}
let buildNameFun: (fname: string, ...rest: string[]) => string = buildName;

this

this是在函數調用時被設定的變量

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();
//出錯,this.suits[pickedSuit] this被綁定到了 window | undefined
alert("card: " + pickedCard.card + " of " + pickedCard.suit);

thisarrow function

  • 通過使用arrow function綁定函數的上下文
  • 可以通過執行加上--noImplicitThis來幫助提醒可能存在的綁定上下文錯誤
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綁定了上下文,但是,this.suits[pickedSuit]的類型依然是any,因為this來源于對象字面量內部的函數表達式
  • 所以TypeScript提供一個this parameter來修復這個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);

這樣就修復了bug,TypeScript知道了this的類型是Deck,就沒有any錯誤了

  • this parameter in callbacks
    當函數作為參數傳入一個一個庫中的函數中的時候,這些庫會把傳入的函數當做普通的函數處理,因此這個函數捕獲的this會變成undefined,所以在這個函數內涉及到this的會報錯。
    通過以下幾步解決錯誤,而且可以使用this
  • 首先通過fake parameter this,確保傳入的參數函數是不會捕獲this,第一層檢查。
interface UIElement {
    addClickListener(onclick: (this: void, e: Event) => void): void;
}
  • 然后通過箭頭函數來幫忙
class Handler {
    info: string;
    onClickGood = (e: Event) => { this.info = e.message }
}

以上可以會有缺點。那就是每次傳建一個Handler類型的對象都會創建一個新的arrow function,造成浪費,解決辦法是在Handler的原型上創建公用的方法


Overloads

JavaScript函數對于不同的參數可能返回不同類型的對象等等,所以需要如下措施來幫助做type-check

  • 通過overloads和調用函數相同名字的函數組,來檢查每一個可能產生的結果。
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);
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