[英]Array of independent generic types in typescript
I'm trying to write a function that will take an array of initialisation functions, along with their options. 我正在尝试编写一个函数,该函数将使用一系列初始化函数及其选项。 I am able derive the type of the options from the function, so I thought I should therefore be able to check that the options I provide are of the same type. 我可以从函数中导出选项的类型,因此我认为我应该能够检查我提供的选项是否具有相同的类型。
The example is a minimised version of this. 该示例是其最小化版本。 I think I need an independent generic for each array item, but I obviously don't know the size of the array. 我认为每个数组项都需要一个独立的泛型,但是我显然不知道数组的大小。
type O<T> = { a: T }
type E<T> = [O<T>, T]
const f = <T={}>(array: E<T>[]) => {}
f([
[{ a: 4 }, 4], //Works
[{ a: "5" }, "5"] //Doesn't work as T is set to number
])
type O = { a: any }
type E<T extends O[]> = {
[P in keyof T]: T[P] extends T[number] ? [T[P], T[P]["a"]] : never;
}
const f = <T extends O[]>(...args: E<T>) => { }
Or more generically as: 或更笼统地说是:
type E<T extends {}[], K extends keyof T[number]> = {
[P in keyof T]: T[P] extends T[number] ? [T[P], T[P][K]] : never;
}
const f = <T extends O[]>(...args: E<T, "a">) => { }
It should be noted that I could not get this to work without using the rest parameter. 应该注意的是,如果不使用rest参数,我将无法正常工作。
Once I stopped focusing so much on a single generic for each element, using mapped types I managed to get to: 一旦我不再对每个元素都只关注一个泛型,就可以使用映射类型来实现:
type E<T extends { a: any }[]> = {
[P in keyof T]: [T[P], T[P]["a"]]
}
But got an error: Type '"a"' cannot be used to index type 'T[P]'. 但是出现错误:类型'“ a”'不能用于索引类型'T [P]'。
This turns out to be because using keyof with tuples includes non-number keys (eg 'length'). 原来是因为对元组使用keyof包含非数字键(例如“长度”)。 I found the solution on a Typescript Github issue ; 我在Typescript Github问题上找到了解决方案; only check the numeric keys. 只检查数字键。
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