TypeScript Functions

TypeScript-Functions

typescript之旅

1.TypeScript-Basic
2.TypeScript interface
3.Typescript-module(1)
4.TypeScript Modules(2)
5.Typescript tsconfig
6.TypeScript Functions
7.Typescript Class

函數

function add(x: number, y: number): number {
    return x + y;
}

let myAdd = function(x: number, y: number): number { return x+y; };

可選參數和默認參數

• typescript會檢查傳遞給一個函數的參數個數與函數指望的參數個數是否一致

function buildName(firstName: string, lastName: string) {
    return firstName + " " + lastName;
}

let result1 = buildName("Bob");                  // error, too few parameters
let result2 = buildName("Bob", "Adams", "Sr.");  // error, too many parameters
let result3 = buildName("Bob", "Adams");         // ah, just right

• 可選參數，參數名後加個?

function buildName(firstName: string, lastName?: string) {
    if (lastName)
        return firstName + " " + lastName;
    else
        return firstName;
}

let result1 = buildName("Bob");  // works correctly now
let result2 = buildName("Bob", "Adams", "Sr.");  // error, too many parameters
let result3 = buildName("Bob", "Adams");  // ah, just right

• 默認參數，參數名後加個=${value}

function buildName(firstName: string, lastName = "Smith") {
    return firstName + " " + lastName;
}

let result1 = buildName("Bob");                  // works correctly now, returns "Bob Smith"
let result2 = buildName("Bob, undefined");       // still works, also returns "Bob Smith"
let result3 = buildName("Bob", "Adams", "Sr.");  // error, too many parameters
let result4 = buildName("Bob", "Adams");         // ah, just right

*注意：可選參數必須位於參數列表末尾，但默認參數能夠不在末尾，用戶必須明確的傳入undefined值來得到默認值

function buildName(firstName = "Will", lastName: string) {
    return firstName + " " + lastName;
}

let result1 = buildName("Bob");                  // error, too few parameters
let result2 = buildName("Bob", "Adams", "Sr.");  // error, too many parameters
let result3 = buildName("Bob", "Adams");         // okay and returns "Bob Adams"
let result4 = buildName(undefined, "Adams");     // okay and returns "Will Adams"

可變參數

function buildName(firstName: string, ...restOfName: string[]) {
  return firstName + " " + restOfName.join(" ");
}

let employeeName = buildName("Joseph", "Samuel", "Lucas", "MacKinzie");

Lambda表達式和使用this

PS：這個有點難度，可是很是重要

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();

alert("card: " + pickedCard.card + " of " + pickedCard.suit);

• 若是咱們運行這個程序，會發現它並無彈出對話框而是報錯了。 由於createCardPicker返回的函數裏的this被設置成了window而不是deck對象。 當你調用cardPicker()時會發生這種狀況。這裏沒有對this進行動態綁定所以爲window。（注意在嚴格模式下，會是undefined而不是window）。

使用lambda表達式(()=>{})能夠解決

let deck = {
    suits: ["hearts", "spades", "clubs", "diamonds"],
    cards: Array(52),
    createCardPicker: function() {
        // Notice: the line below is now a lambda, allowing us to capture `this` earlier
        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);

重載

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);

這樣改變後，重載的pickCard函數在調用的時候會進行正確的類型檢查。

注意，function pickCard(x): any並非重載列表的一部分，所以這裏只有兩個重載：一個是接收對象另外一個接收數字。 以其它參數調用pickCard會產生錯誤。