为什么我不能返回通用的"T"来满足Partial< T&gt ;? [英] Why can't I return a generic 'T' to satisfy a Partial<T>?

问题描述

我用TypeScript编写了一些代码:

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 };
}

这会产生错误:

Type '{ x: 10; }' is not assignable to type 'Partial<T>'

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

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，要记住一个重要规则

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

您为getThing ...

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

...暗示着这样的合法调用，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 } 是是合法的Partial<Point3D>.

但是子类型化功能不仅仅适用于添加其他属性-子类型化可以包括选择属性本身域的一组更受限的集合.您可能具有这样的类型:

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 };

现在编写时

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

p.x的值为10，不是合法的UnitPoint.

如果您发现自己处于这种情况，则很可能您的返回类型实际上不是通用的.更准确的功能签名应该是

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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