如何在 TypeScript 中声明函数的返回类型
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How to declare Return Types for Functions in TypeScript
提问by Tarik
I checked here https://github.com/Microsoft/TypeScript/blob/master/doc/spec.mdwhich is the TypeScript Language Specificationsbut I couldn't see one thing that how I can declare a return type of the function. I showed what I was expecting in the code below : greet(name:string) :string {}
我在这里查看了https://github.com/Microsoft/TypeScript/blob/master/doc/spec.md，这是TypeScript 语言规范，但我看不到如何声明函数的返回类型的一件事。我在下面的代码中展示了我的期望：greet(name:string) :string {}
class Greeter {
greeting: string;
constructor (message: string) {
this.greeting = message;
}
greet() : string{
return "Hello, " + this.greeting;
}
}
I see we can use something (name:string) => anybut they are used mostly when passing callback functions around:
我看到我们可以使用一些东西，(name:string) => any但它们主要在传递回调函数时使用：
function vote(candidate: string, callback: (result: string) => any) {
// ...
}
采纳答案by Fenton
You are correct - here is a fully working example - you'll see that var resultis implicitly a string because the return type is specified on the greet()function. Change the type to numberand you'll get warnings.
你是对的 - 这是一个完全有效的例子 - 你会看到它var result隐式是一个字符串，因为返回类型是在greet()函数上指定的。将类型更改为number，您将收到警告。
class Greeter {
greeting: string;
constructor (message: string) {
this.greeting = message;
}
greet() : string {
return "Hello, " + this.greeting;
}
}
var greeter = new Greeter("Hi");
var result = greeter.greet();
Here is the number example - you'll see red squiggles in the playground editor if you try this:
这是数字示例 - 如果您尝试这样做，您将在操场编辑器中看到红色波浪线：
greet() : number {
return "Hello, " + this.greeting;
}
回答by mohamed hegazy
You can read more about function types in the language specificationin sections 3.5.3.5 and 3.5.5.
您可以在 3.5.3.5 和 3.5.5 节的语言规范中阅读有关函数类型的更多信息。
The TypeScript compiler will infer types when it can, and this is done you do not need to specify explicit types. so for the greeter example, greet() returns a string literal, which tells the compiler that the type of the function is a string, and no need to specify a type. so for instance in this sample, I have the greeter class with a greet method that returns a string, and a variable that is assigned to number literal. the compiler will infer both types and you will get an error if you try to assign a string to a number.
TypeScript 编译器会在可能的情况下推断类型，这样做后您不需要指定显式类型。所以对于greeter的例子，greet()返回一个字符串字面量，它告诉编译器函数的类型是一个字符串，不需要指定类型。因此，例如在这个示例中，我有一个带有greet 方法的greeter 类，该方法返回一个字符串，以及一个分配给数字文字的变量。编译器将推断这两种类型，如果您尝试将字符串分配给数字，则会出现错误。
class Greeter {
greet() {
return "Hello, "; // type infered to be string
}
}
var x = 0; // type infered to be number
// now if you try to do this, you will get an error for incompatable types
x = new Greeter().greet();
Similarly, this sample will cause an error as the compiler, given the information, has no way to decide the type, and this will be a place where you have to have an explicit return type.
同样，这个示例将导致错误，因为编译器在给定信息的情况下无法确定类型，这将是您必须具有显式返回类型的地方。
function foo(){
if (true)
return "string";
else
return 0;
}
This, however, will work:
但是，这将起作用：
function foo() : any{
if (true)
return "string";
else
return 0;
}
回答by Luca C.
functionName() : ReturnType { ... }
回答by Adverbly
Return types using arrow notation is the same as previous answers:
使用箭头符号的返回类型与之前的答案相同：
const sum = (a: number, b: number) : number => a + b;
