[TypeScript][文档]函数

在JavaScript里,函数可以使用函数体外部的变量。 当函数这么做时,我们说它‘捕获’了这些变量。

let z = 100;
function addToZ(x, y) {
    return x + y + z;
}

一、函数类型

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

myAdd 具有 函数类型   (x: number, y: number) => number

匿名函数的参数返回值也分别具有类型  function(x: number, y: number): number

【注】:推断类型——可以不写左边或者右边的类型,TS会自动推断出来,如:

// myAdd has the full function type
let myAdd = function(x: number, y: number): number { return x + y; };

// The parameters `x` and `y` have the type number
let myAdd: (baseValue: number, increment: number) => number =
    function(x, y) { return x + y; };

二、函数参数

——传递给一个函数的参数个数必须与函数期望的参数个数一致。

1.可选参数:

参数可传可不传。 没传参的时候,它的值就是undefined。

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

【注】:可选参数必须跟在必须参数后面

2.默认参数

当用户没有传递这个参数或传递的值是undefined时,可以为参数提供一个默认值。

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】:默认参数都是可选的,因此可选参数与默认参数 共享参数类型:

(firstName: string, lastName?: string) => string

【注】:带默认值的参数不需要放在必须参数的后面。 如果带默认值的参数出现在必须参数前面,用户必须明确的传入 undefined来获得默认值。

3.剩余参数

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

let buildNameFun: (fname: string, ...rest: string[]) => string = buildName;

三、this

this的值在函数被调用的时候才会指定。 因此需要知道函数调用的上下文是什么。

1.箭头函数

箭头函数能保存函数创建时的 this值,而不是调用时的值:

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);

【疑】this参数、this回调函数里的参数

四、重载

根据传给函数不同的参数,获取不同的处理

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 { // 这个不是指参数为any类型的重载,而是前两个重载的实现
    // 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);

【注】:function pickCard(x): any并不是重载列表的一部分,因此这里只有两个重载:一个是接收对象另一个接收数字。 以其它参数调用 pickCard会产生错误。

 

posted @ 2018-12-14 14:16  麦田里的小王子  阅读(176)  评论(0编辑  收藏  举报