[英]TypeScript - function argument types equality
I have a problem with using a generic in a class.我在 class 中使用泛型时遇到问题。 There's a class Item
that accepts a generic T
that extends constraining ItemValue
type and there's a type Items
that contains instances of Item
typed as Item<ItemValue
>.有一个 class Item
接受一个扩展约束ItemValue
类型的通用T
,并且有一个类型Items
包含类型为Item<ItemValue
> 的Item
实例。
The problem is that the compiler does not pass a potentially valid generic argument through ItemValue
constraint because callback
property of Item
is using it as a callback value.问题是编译器没有通过ItemValue
约束传递潜在有效的泛型参数,因为Item
的callback
属性将其用作回调值。
If I removed the callback
property from the class, there would be no error.如果我从 class 中删除callback
属性,则不会出现错误。
Further, I tried to replicate this without class, and there's no such error.此外,我尝试在没有 class 的情况下复制它,并且没有此类错误。 Do you think it's a bug, or am I doing something wrong?你认为这是一个错误,还是我做错了什么?
TypeScript Playground TypeScript 游乐场
type ItemValue = number | string;
type Callback<T> = (value: T) => void;
class Item<T extends ItemValue> {
private value: T;
// If you comment this out, there will be no error
private callback: Callback<T> = () => { }
constructor(value: T) {
this.value = value;
}
}
type Items = {
[index: string]: Item<ItemValue>
};
let items: Items = {
a: new Item<string>('') // error
}
// Type 'Item<string>' is not assignable to type 'Item<ItemValue>'.
// Types of property 'callback' are incompatible.
// Type 'Callback<string>' is not assignable to type 'Callback<ItemValue>'.
// Type 'ItemValue' is not assignable to type 'string'.
// Type 'number' is not assignable to type 'string'.
// This will also result in a similar error
// type AnItem = Item<ItemValue>;
// let a: AnItem = new Item<string>('');
// Here trying to do the same without a class.
function addCallback<T extends ItemValue>(cb: Callback<T>) {
}
function prepareAddCallback(cb: Callback<string>) {
// No problems with putting Callback<string> to Callback<ItemValue>
// Is it a bug?
addCallback(cb);
}
As @Joey pointed out in his answer:正如@Joey 在他的回答中指出的那样:
string
is assignable tostring | number
string
可分配给string | number
string | number
, but(val: string) => void
is not assignable to(val: string | number) => void
string | number
,但是(val: string) => void
不能分配给(val: string | number) => void
type CallbackStrOrNum = (value: string | number) => void;
const addCallback = (cb: CallbackStrOrNum) => { };
addCallback((value: string) => { }); // error
let x: string | number = ''; // ok
This narrows down the issue.这缩小了问题的范围。 My question now would be, why is that?我现在的问题是,为什么会这样?
To be clear, I solved the issue in this example by defining private callback: Callback<ItemValue> = () => { }
instead of private callback: Callback<T> = () => { }
Having the generic in the callback parameter was not critical here.为了清楚起见,我通过定义private callback: Callback<ItemValue> = () => { }
而不是private callback: Callback<T> = () => { }
在回调参数中有泛型在这里并不重要。 The lesson here is to not define function types properties that have a generic, as then the class will be inextensible, eg Item<number>
would not be assignable to Item<string | number>
这里的教训是不要定义具有泛型的 function 类型属性,因为这样 class 将是不可扩展的,例如Item<number>
不能分配给Item<string | number>
Item<string | number>
Also, in the real example, it was a dictionary of callbacks, in case you are wondering why I didn't define callback
as a regular method. Item<string | number>
此外,在实际示例中,它是一个回调字典,以防您想知道为什么我没有将callback
定义为常规方法。
Note that string
is assignable to string | number
请注意, string
可分配给string | number
string | number
, but (val: string) => void
is not assignable to (val: string | number) => void
. string | number
,但是(val: string) => void
不能分配给(val: string | number) => void
。
In your example, Item<string>
is not actually assignable to Item<string | number>
在您的示例中, Item<string>
实际上不能分配给Item<string | number>
Item<string | number>
because the callback
properties don't have compatible types (ie (value: string) => void
is not assignable to (value: string | number) => void
). Item<string | number>
因为callback
属性没有兼容的类型(即(value: string) => void
不能分配给(value: string | number) => void
)。
Edit:编辑:
To be clear, this is by design and is a natural consequence of the type union operator ( |
).需要明确的是,这是设计使然,并且是类型联合运算符 ( |
) 的自然结果。 Otherwise, the following code would compile, which would certainly be an issue.否则,下面的代码会编译,这肯定是个问题。
const callback: (value: string | number) => void = (str: string) => {str.toLowerCase()};
callback(4);
I changed this and it removed the error:我改变了这个,它消除了错误:
let items: Items = {
a: new Item<ItemValue>('') // (No) error
}
I believe it's because ItemValue can be a string OR a number so Item< string> wouldn't specific enough.我相信这是因为 ItemValue 可以是字符串或数字,所以 Item< string> 不够具体。 It would also work with:它也适用于:
let items: Items = {
a: new Item<string | number>('') // (No) error
}
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