Typescript 在这种情况下，为什么不从参数推断泛型值N？

这个问题：

询问如何创建需要两个长度相同的数组的函数

这是我试图解决的问题

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. 

type arrayOfficedLength=readonly T[]和{length:N}；
常数a1:ArrayOfficedLength=[1]作为常数//预期误差
常数a2:ArrayOfficedLength=[1,2]作为常数；
函数myFunction（array1:ArrayOfficedLength

如前所述，如果您显式地声明了泛型值
N
——arrray的长度，则此代码仅给出TypeScript错误

为什么TypeScript无法从传递到函数中的参数推断出值N？
您需要提示编译器期望元组类型。否则编译器会将数组文本（如
[2,3,4]
扩展到
数字[]
。提示的形式通常是在类型注释或泛型约束中包含一个元组类型；最好是某个元组类型的并集，它不会妨碍您的操作：

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

我觉得不错。好吧，希望有帮助，祝你好运

Ugh.我越来越意识到，虽然typescript可能可以做你想做的一切，但要做到这一点很难。
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