打字稿在使用箭头函数“原型”时不会出错

Question

这段代码：

let arrowFn = () => {};
arrowFn.prototype.blah;

在编译时抛出一个错误，因为箭头函数没有 prototypes 但 typescript 没有错误并愉快地转换代码。

为什么？

Answer 1

这里你已经内置了 FUnction 类型：

interface Function {
    /**
     * Calls the function, substituting the specified object for the this value of the function, and the specified array for the arguments of the function.
     * @param thisArg The object to be used as the this object.
     * @param argArray A set of arguments to be passed to the function.
     */
    apply(this: Function, thisArg: any, argArray?: any): any;

    /**
     * Calls a method of an object, substituting another object for the current object.
     * @param thisArg The object to be used as the current object.
     * @param argArray A list of arguments to be passed to the method.
     */
    call(this: Function, thisArg: any, ...argArray: any[]): any;

    /**
     * For a given function, creates a bound function that has the same body as the original function.
     * The this object of the bound function is associated with the specified object, and has the specified initial parameters.
     * @param thisArg An object to which the this keyword can refer inside the new function.
     * @param argArray A list of arguments to be passed to the new function.
     */
    bind(this: Function, thisArg: any, ...argArray: any[]): any;

    /** Returns a string representation of a function. */
    toString(): string;

    prototype: any;
    readonly length: number;

    // Non-standard extensions
    arguments: any;
    caller: Function;
}

如您所见，FUnction.prototype 具有 any 类型。这就是为什么你没有任何错误。

let arrowFn = () => {};

function x(){}


x.prototype
arrowFn.prototype

x 和 arrowFn 都指向同一个 Function.prototype 类型。

如果您将箭头函数替换为function allowFn(){}，您将只收到undefined，不会出错

更新

const x = () => { }
const y = function () { }
function z() { }

type X = typeof x // () => void
type Y = typeof y // () => void
type Z = typeof z // () => void