在 TypeScript 中可以将强类型函数作为参数吗? [英] Are strongly-typed functions as parameters possible in TypeScript?
问题描述
在 TypeScript 中,我可以将函数的参数声明为函数类型.是否有一种我缺少的类型安全"的方法来做到这一点?例如,考虑这个:
In TypeScript, I can declare a parameter of a function as a type Function. Is there a "type-safe" way of doing this that I am missing? For example, consider this:
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);
保存回调不是类型安全的,我给它一个回调函数,其中函数的参数是一个字符串,但我传递了一个数字,并且编译没有错误.我可以在保存类型安全的函数中设置结果参数吗?
The save callback is not type safe, I am giving it a callback function where the function's parameter is a string but I am passing it a number, and compiles with no errors. Can I make the result parameter in save a type-safe function?
TL;DR 版本:在 TypeScript 中是否有等效的 .NET 委托?
TL;DR version: is there an equivalent of a .NET delegate in TypeScript?
推荐答案
好的.一个函数的 type 由它的类型组成参数及其返回类型.这里我们指定callback
参数的类型必须是接受数字并返回类型any
的函数":
Sure. A function's type consists of the types of its argument and its return type. Here we specify that the callback
parameter's type must be "function that accepts a number and returns type any
":
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
如果需要,您可以定义一个类型别名 封装这个:
If you want, you can define a type alias to encapsulate this:
type NumberCallback = (n: number) => any;
class Foo {
// Equivalent
save(callback: NumberCallback) : void {
callback(42);
}
}
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