为什么在这种情况下不能从参数中推导出泛型值N? [英] Why is the generic value N not inferred from arguments in this case?

问题描述

这个问题:

TypeScript:要求两个数组的长度相同吗?

询问如何创建需要两个相同长度数组的函数.

这是我尝试解决的方法.

type ArrayOfFixedLength<T extends any, N extends number> = readonly T[] & { length: N }; 

const a1: ArrayOfFixedLength<number, 2> = [1] as const; //expected error
const a2: ArrayOfFixedLength<number, 2> = [1, 2] as const; 


function myFunction<N extends number>(array1: ArrayOfFixedLength<any, N >, array2: ArrayOfFixedLength<any, N>) {
return true; 
}

myFunction<3>([1, 2, 3] as const, [2, 3, 4] as const); 
myFunction<2>([1, 2] as const, [1, 2, 3] as const); //expected error

// However, if you don't specify the array length, 
// It fails to error
myFunction([1, 2, 3] as const, [2, 3, 4] as const); 
myFunction([1, 2] as const, [1, 2, 3] as const); // error is expected, but there is none. 

解决方案

您需要给编译器一个提示，以期望使用元组类型.否则，编译器会将像[2, 3, 4]这样的数组文字扩展为number[].提示通常采取在类型注释或通用约束中包括元组类型的形式.最好是一些元组类型的并集，它不会妨碍您的工作:

function myFunction<N extends number>(
    array1: ArrayOfFixedLength<any, N> | [never],
    array2: ArrayOfFixedLength<any, N & {}> | [never]) {
    return true;
}

| [never]是一个提示.请参见 microsoft/TypeScript#27179 ，尤其是 microsoft/TypeScript#30680 中找到一种更简单的方法，但我不知道不知道这样的事情是否会付诸实施.

---

此外，请注意，对于array2，我已将N替换为N & {}.如果不这样做，则函数将无法将两个数组限制为相同的长度.而是将N推断为两个数组的长度的并集.理想情况下，您只希望 array1用于推断 N，并且array2中的N应该是非推断"的.有一个公开的GitHub问题， microsoft/TypeScript#14829 ，要求支持这一点. & {}是降低推理优先级的技术，足以仅从array1推断N而不是array2.

让我们看看它的工作原理:

myFunction([1, 2, 3] as const, [2, 3, 4] as const); // okay
myFunction([1, 2] as const, [1, 2, 3] as const); // error
myFunction([1, 2, 3], [2, 3, 4]); // okay
myFunction([1, 2], [1, 2, 3]); // error

对我很好.好吧，希望能有所帮助；祝你好运！

游乐场连结代码

This question:

TypeScript: Require that two arrays be the same length?

Asks how to create a function that requires two arrays of the same length.

Here's my attempt at a solution.

type ArrayOfFixedLength<T extends any, N extends number> = readonly T[] & { length: N }; 

const a1: ArrayOfFixedLength<number, 2> = [1] as const; //expected error
const a2: ArrayOfFixedLength<number, 2> = [1, 2] as const; 


function myFunction<N extends number>(array1: ArrayOfFixedLength<any, N >, array2: ArrayOfFixedLength<any, N>) {
return true; 
}

myFunction<3>([1, 2, 3] as const, [2, 3, 4] as const); 
myFunction<2>([1, 2] as const, [1, 2, 3] as const); //expected error

// However, if you don't specify the array length, 
// It fails to error
myFunction([1, 2, 3] as const, [2, 3, 4] as const); 
myFunction([1, 2] as const, [1, 2, 3] as const); // error is expected, but there is none. 

Playground

As noted, this code only gives the TypeScript error if you explicitly state the generic value N - the length of the arrray.

Why is TypeScript unable to infer the value N from the arguments passed into the function?

解决方案

You need to give a hint to the compiler to expect a tuple type. Otherwise the compiler will widen an array literal like [2, 3, 4] to number[]. The hint usually takes the form of including a tuple type in the type annotation or generic constraint; preferably a union of some tuple type that doesn't get in the way of what you're doing:

function myFunction<N extends number>(
    array1: ArrayOfFixedLength<any, N> | [never],
    array2: ArrayOfFixedLength<any, N & {}> | [never]) {
    return true;
}

The | [never] is a hint. See microsoft/TypeScript#27179 and specifically this comment for more information. I used [never] since I expect that you won't pass any arrays with a never value, so it won't matter in practice if array1 accepts such an array.

Yes, it's ugly. I've asked for an easier way in microsoft/TypeScript#30680, but I don't know if something like that will ever be implemented.

---

Also, note that for array2 I've replaced N with N & {}. If you don't do that, your function will not serve the purpose of constraining the two arrays to the same length; instead, N will just be inferred as the union of the lengths of the two arrays. Ideally you want only array1 to be used to infer N, and the N in array2 should be "non-inferential". There's an open GitHub issue, microsoft/TypeScript#14829, asking for support for this. The & {} is a technique to lower the inference priority, which suffices to only infer N from array1 and not array2.

Let's see it work:

myFunction([1, 2, 3] as const, [2, 3, 4] as const); // okay
myFunction([1, 2] as const, [1, 2, 3] as const); // error
myFunction([1, 2, 3], [2, 3, 4]); // okay
myFunction([1, 2], [1, 2, 3]); // error

Looks good to me. Okay, hope that helps; good luck!

Playground link to code

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