为什么我不能返回一个通用的“T"来满足 Partial<T>? [英] Why can't I return a generic 'T' to satisfy a Partial<T>?

问题描述

我用 TypeScript 写了一些代码:

I wrote some code in TypeScript:

type Point = {
  x: number;
  y: number;
};
function getThing<T extends Point>(p: T): Partial<T> {
  // More interesting code elided
  return { x: 10 };
}

这会产生一个错误:

输入 '{ x: 10;}' 不可分配给类型 'Partial'

这似乎是一个错误 - { x: 10 } 显然是一个 Partial.TypeScript 在这里做错了什么?我该如何解决这个问题?

This seems like a bug - { x: 10 } is clearly a Partial<Point>. What's TypeScript doing wrong here? How do I fix this?

推荐答案

在考虑编写泛型函数时，有一个重要的规则要记住

When thinking about writing a generic function, there's an important rule to remember

您为 getThing 提供的合约 ...

The contract you've provided for getThing ...

function getThing<T extends Point>(p: T): Partial<T>

... 暗示像这样的合法调用，其中 T 是 Point 的子类型:

... implies legal invocations like this one, where T is a subtype of Point:

const p: Partial<Point3D> = getThing<Point3D>({x: 1, y: 2, z: 3});

当然，{ x: 10 } 是一个合法的Partial.

但是子类型化的能力不仅仅适用于添加额外的属性——子类型化可以包括选择一组更受限制的属性本身的域.你可能有这样的类型:

But the ability to subtype doesn't just apply to adding additional properties -- subtyping can include choosing a more restricted set of the domain of the properties themselves. You might have a type like this:

type UnitPoint = { x: 0 | 1, y: 0 | 1 };

现在当你写

const p: UnitPoint = getThing<UnitPoint>({ x: 0, y: 1});

p.x 的值为 10，它不是一个合法的 UnitPoint.

p.x has the value 10, which is not a legal UnitPoint.

如果您发现自己处于这样的情况，很有可能您的返回类型实际上不是通用的.更准确的函数签名是

If you find yourself in a situation like this, odds are good that your return type is not actually generic. A more accurate function signature would be

function getThing<T extends Point>(p: T): Partial<Point> {

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