如何在 Typescript 函数接口中约束泛型类型? [英] How do you constrain a Generic type in a Typescript Function Interface?

问题描述

假设我们有以下接口:

interface SomeInterface {a: string; b: number}
interface SomeFunction<T> {(arg: T) :T}

我们可以像这样使用函数接口:

We can use the function interface like so:

const myFun: SomeFunction<string> = arg => arg;

但问题是我们将泛型类型 T 指定为字符串，因此它不再是泛型.我想保持类型 T 通用但也限制它.

but the issue with that is we specify the generic type T as string, so it is no longer generic. I would like to keep the type T generic but also constrain it.

声明 SomeFunction 如:

interface SomeFunction {<T>(arg: T) : T}

允许它像一般使用:

const myFun: SomeFunction = (arg) => arg;

但是我们失去了在函数声明中约束 T 的机会.

but then we lose the chance to constrain T at the function declaration.

我想要实现 SomeFunction 的函数来定义 T 必须扩展 SomeInterface 并且我想要另一个也实现 SomeFunction 来定义 T 扩展其他东西.

I would like my function that implements SomeFunction to define that T must extend SomeInterface and I would like another function that also implements SomeFunction to define that T extends something else.

所以，基本上，我想要这样的东西:

So, basically, I want something like this:

function myFun2<T extends SomeInterface>(arg: T): T {
    console.log(arg.a);
    return arg;
}

function myFun3<T extends {c: string }>(arg: T): T {
    console.log(arg.c);
    return arg;
}

我在上述解决方案中遇到的问题是这些声明中没有提到 SomeFunction 接口，即使我们基本上匹配它.

The issue I have with the solution above is that there is no mention of the SomeFunction interface in these declarations even though we basically match it.

是否有更好的方法来声明 myFun2 和 myFun3 以便我们确保它们符合 SomeFunction?

Is there a better way to declare myFun2 and myFun3so we make sure they conform to SomeFunction?

推荐答案

如果您希望能够在调用站点定义约束，您可以使用约束类型来概括 SomeFunction，例如

If you want the ability to define constraints at call site, you can generalise SomeFunction with a constraint type, like

interface SomeFunction<C> {
    <T extends C>(arg: T) : T
}

并在定义函数时提供正确的约束

and provide the correct constraint when defining your functions

const myFun2: SomeFunction<SomeInterface> = arg => {
    console.log(arg.a) // ok
    return arg
}

const myFun3: SomeFunction<{ c: string }> = arg => {
    console.log(arg.c) // ok
    return arg
}

如果你想允许不受约束的函数，你也可以给一个 unknown 类型的默认约束，比如

You could also give a default constraint of type unknown if you wanted to allow unconstrained functions, like

interface SomeFunction<C = unknown> {
    <T extends C>(arg: T): T
}

const unconstrained: SomeFunction = arg => arg

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