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

函数

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

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

可选参数和默认参数

• typescript会检查传递给一个函数的参数个数与函数指望的参数个数是否一致

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

• 可选参数，参数名后加个?

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

• 默认参数，参数名后加个=${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：可选参数必须位于参数列表末尾，但默认参数能够不在末尾，用户必须明确的传入undefined值来得到默认值

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"

可变参数

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

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

Lambda表达式和使用this

PS：这个有点难度，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);

• 若是咱们运行这个程序，会发现它并无弹出对话框而是报错了。 由于createCardPicker返回的函数里的this被设置成了window而不是deck对象。 当你调用cardPicker()时会发生这种状况。这里没有对this进行动态绑定所以为window。（注意在严格模式下，会是undefined而不是window）。

使用lambda表达式(()=>{})能够解决

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);

重载

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);

这样改变后，重载的pickCard函数在调用的时候会进行正确的类型检查。

注意，function pickCard(x): any并非重载列表的一部分，所以这里只有两个重载：一个是接收对象另外一个接收数字。 以其它参数调用pickCard会产生错误。