[英]How to infer subtype of generic in TypeScript?
I have a function with some argument's type:我有一个带有一些参数类型的 function:
function func<T extends {field1: string}>(arg1: T) {
// function's code
}
I have a familiar interface with generic type, like this:我有一个熟悉的泛型接口,如下所示:
interface SomeInterface<T> {
field1: string;
}
Now i want to infer generic parameter of interface and assign it to function's return type:现在我想推断接口的通用参数并将其分配给函数的返回类型:
const value1: SomeInterface<number> = {field1: "Hello"};
const res1 = func(value1);
I need a type for res1 is number .我需要res1的类型是number 。
What a function's signature i need to use?我需要使用什么函数签名?
function func<T extends {field1: string}>(arg1: T) : T extends ???? { // What?
There is no point in using a generic type in an interface if the type is not being referenced by any attribute.如果类型没有被任何属性引用,那么在接口中使用泛型类型是没有意义的。
Here is an interface which receives a generic type and applies that type to one of its attributes:这是一个接收泛型类型并将该类型应用于其属性之一的接口:
interface SomeInterface<T> {
field1: T;
}
In other hand, if you want to create function with a generic return type:另一方面,如果要创建具有通用返回类型的 function:
function func<T>(): T {
// code
}
// function call
const res1 = func<number>();
And finally, if you want your function to receive as argument a generic type and return another generic type:最后,如果您希望您的 function 接收一个泛型类型作为参数并返回另一个泛型类型:
function func<T, R>(arg: T) : R {
// code
}
// function call
const res1 = func<SomeInterface, number>(value1);
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