打字稿中的通用铸造 [英] Generic casting in Typescript

问题描述

我试图在TypeScript中创建一个简单的通用查找函数，如下所示：

I'm trying to create a simple generic find function in TypeScript, along the lines of:

export function findFirst<T, U>(
        array: T[], 
        predicate: (item: T) => boolean, 
        selector?: (item: T) => U): U {
     ...
}

所以，我的参数是：

- 要过滤的数组

- 用于测试每个元素的谓词

- 用于获取返回值的选择器

So, my parameters are:
- the array to filter through
- a predicate to test each element
- a selector to get the return value

我想要做的是提供一个默认选择器，即如果没有提供选择器，只返回整个值，例如：

What I want to do is provide a default selector, i.e. if no selector is provided, just return the whole value, i.e. something like:

if (typeof selector === "undefined")
    selector = (x) => x;

但是，这样做（或者甚至（x）=>< U> ; x ）打破了函数的通用定义。如何在不删除通用参数的情况下实现默认选择器？

However this, (or even (x) => <U>x) breaks the generic definition of the function. How can I achieve a default selector without removing the generic parameters?

如果我使用如下代码：

var arr = [1,2,3,4];
var even = findFirst(arr, x => x % 2 === 0);

返回第一个偶数，它推导出y的类型为 {} ，即 object 而不是数字。
看起来，因为U只能从选择器参数中推断，在这种情况下是未定义的，U默认为 object 。
我知道我有太多的类型推理，但是有什么办法呢？

i.e. return the first even number, it infers the type of y as {} i.e. object instead of number. It seems that, as U can only be inferred from the selector parameter, which is undefined in this case, U defaults to object. I know I ask a bit too much of the type inference, but is there any way around this?

推荐答案

这里是完整的代码：

export function findFirst<T, U>(
        array: T[], 
        predicate: (item: T) => boolean, 
        selector: (item: T) => U = (x:T)=> <U><any>x): U {
     return array.filter(predicate).map(selector)[0];
}

的原因< U>< any> ：只能键入 U 键入 T 如果T是U的子类型或U是T的子类型。因为这不是可确定的，所以需要在之前转换为< any> 您可以断言< U>

Reason for <U><any> : Type T can be asserted to type U only if T is a subtype of U OR U is a subtype of T. Since that is not determinable you need to convert to <any> before you can assert to <U>

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