在TypeScript中，我可以将函数的形参声明为function类型。是否有一种“类型安全”的方式来做到这一点，我错过了?例如，考虑这个:
class Foo {
save(callback: Function) : void {
//Do the save
var result : number = 42; //We get a number from the save operation
//Can I at compile-time ensure the callback accepts a single parameter of type number somehow?
callback(result);
}
}
var foo = new Foo();
var callback = (result: string) : void => {
alert(result);
}
foo.save(callback);
保存回调是不类型安全的，我给它一个回调函数，其中函数的参数是一个字符串，但我传递给它一个数字，并编译没有错误。我可以使结果参数在保存类型安全的函数?
TL;DR版本:在TypeScript中有等价的。net委托吗?
我知道这篇文章很旧了，但有一个更紧凑的方法，与所要求的略有不同，但可能是一个非常有用的替代方案。你可以在调用方法时内联声明这个函数(在本例中是Foo的save())。它看起来是这样的:
class Foo {
save(callback: (n: number) => any) : void {
callback(42)
}
multipleCallbacks(firstCallback: (s: string) => void, secondCallback: (b: boolean) => boolean): void {
firstCallback("hello world")
let result: boolean = secondCallback(true)
console.log("Resulting boolean: " + result)
}
}
var foo = new Foo()
// Single callback example.
// Just like with @RyanCavanaugh's approach, ensure the parameter(s) and return
// types match the declared types above in the `save()` method definition.
foo.save((newNumber: number) => {
console.log("Some number: " + newNumber)
// This is optional, since "any" is the declared return type.
return newNumber
})
// Multiple callbacks example.
// Each call is on a separate line for clarity.
// Note that `firstCallback()` has a void return type, while the second is boolean.
foo.multipleCallbacks(
(s: string) => {
console.log("Some string: " + s)
},
(b: boolean) => {
console.log("Some boolean: " + b)
let result = b && false
return result
}
)
对于可能成功或失败的网络调用之类的事情，multipleCallback()方法非常有用。再次假设一个网络调用示例，当调用multipleCallbacks()时，成功和失败的行为都可以在一个地方定义，这有助于将来的代码读者更加清晰。
一般来说，根据我的经验，这种方法更简洁，更少混乱，整体上更清晰。
祝大家好运!
当然。函数的类型由其实参类型和返回类型组成。这里我们指定回调参数的类型必须是“接受数字并返回any类型的函数”:
class Foo {
save(callback: (n: number) => any) : void {
callback(42);
}
}
var foo = new Foo();
var strCallback = (result: string) : void => {
alert(result);
}
var numCallback = (result: number) : void => {
alert(result.toString());
}
foo.save(strCallback); // not OK
foo.save(numCallback); // OK
如果你愿意，你可以定义一个类型别名来封装它:
type NumberCallback = (n: number) => any;
class Foo {
// Equivalent
save(callback: NumberCallback) : void {
callback(42);
}
}
因为不能轻易地将函数定义和另一种数据类型结合起来，所以我发现使用这些类型对强类型很有用。根据德鲁的回答。
type Func<TArgs extends any[], TResult> = (...args: TArgs) => TResult;
//Syntax sugar
type Action<TArgs extends any[]> = Func<TArgs, undefined>;
现在您可以强类型每个参数和返回类型!下面是一个比上面的参数更多的例子。
save(callback: Func<[string, Object, boolean], number>): number
{
let str = "";
let obj = {};
let bool = true;
let result: number = callback(str, obj, bool);
return result;
}
现在，您可以编写联合类型，如对象或返回对象的函数，而无需创建可能需要导出或使用的全新类型。
//THIS DOESN'T WORK
let myVar1: boolean | (parameters: object) => boolean;
//This works, but requires a type be defined each time
type myBoolFunc = (parameters: object) => boolean;
let myVar1: boolean | myBoolFunc;
//This works, with a generic type that can be used anywhere
let myVar2: boolean | Func<[object], boolean>;
function callbackTesting(callbacks: {onYes: (data: any) => void,onNo: (data: any) => void,onError: (err: any) => void,}, type: String){
switch(type){
case "one":
callbacks.onYes("Print yes");
break;
case "two":
callbacks.onNo("Print no");
break;
default:
callbacks.onError("Print error");
break;
}
}
const onYes1 = (data: any) : void => {
console.log(data);
}
const onNo1 = (data: any) : void => {
console.log(data);
}
const onError1 = (data: any) : void => {
console.log(data);
}
callbackTesting({onYes: function (data: any) {onYes1(data);},onNo: function (data: any) {onNo1(data);},onError: function (data: any) {onError1(data);}}, "one");
callbackTesting({onYes: function (data: any) {onYes1(data);},onNo: function (data: any) {onNo1(data);},onError: function (data: any) {onError1(data);}}, "two");
callbackTesting({onYes: function (data: any) {onYes1(data);},onNo: function (data: any) {onNo1(data);},onError: function (data: any) {onError1(data);}}, "cfhvgjbhkjlkm");
