TypeScript需要提供通用参数 [英] TypeScript require generic parameter to be provided

问题描述

我具有以下功能:

async function get<U>(url: string): Promise<U> {
    return getUrl<u>(url);
}

但是，可以这样称呼它(TS将U设置为any):

However, it is possible to call it like this (U is set to any by TS):

get('/user-url');

有没有一种方法可以定义此函数，使得它需要显式提供U，如

Is there a way to define this function such that it requires U to be provided explicitly, as in

get<User>('/user-url');

推荐答案

对此没有内置支持，但是我们可以设计一个方案，其中不传入类型参数将使用默认的泛型类型参数生成错误.和条件类型.即，我们将给 U 一个默认值 void .如果默认值是 U 的实际值，那么我们将在函数中键入参数，而该参数实际上不应传递给它，以获取错误:

There is no built-in support for this, we can however engineer a scenario where not passing in a type parameter will generate an error, using default generic type arguments and conditional types. Namely we will give U a default value of void. If the default value is the actual value of U, then we will type the parameter to the function as something that should not really be passed in so as to get an error:

async function get<U = void>(url: string & (U extends void ? "You must provide a type parameter" : string)): Promise<U> {
    return null as any;
}

get('/user-url'); // Error Argument of type '"/user-url"' is not assignable to parameter of type '"You must provide a type parameter"'.

class User {}
get<User>('/user-url');

错误消息不是理想的，但是我认为它将使消息得到传播.

The error message is not ideal, but I think it will get the message across.

编辑:有关在参数类型中使用类型参数的解决方案，请参见

Edit: For a solution where the type parameter is used in parameter types see here

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