如何为 Typescript 接口中的函数定义 void 返回类型？
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How can I define a return type of void for a function in a Typescript interface?
提问by Alan2
I have this function:
我有这个功能：
network = (action): void =>{
if (action) {
this.action = action;
this.net = true;
this.netd = true;
} else {
this.action = null;
this.net = false;
this.netd = false;
}
}
I tried to define an interface but it's not working for me:
我试图定义一个接口，但它对我不起作用：
interface IStateService {
network: (action: string): void;
}
I get a message saying "unexpected token" on void
我在 void 上收到一条消息，说“意外的令牌”
回答by Ryan Cavanaugh
You have two options for syntax for a function-typed interface member, which are equivalent here:
函数类型接口成员的语法有两个选项，它们在此处等效：
interface IStateService {
network: (action: string) => void;
}
or
或者
interface IStateService {
network(action: string): void;
}
回答by Douglas
That is close to the "type literal" syntax, but the braces are required for that:
这接近于“类型文字”语法，但需要大括号：
interface IStateService {
network: { (action: string): void; }
}
This is the full syntax, and allows defining overloads, like this:
这是完整的语法，并允许定义重载，如下所示：
interface IStateService {
network: {
(): string;
(action: string): void;
}
}
