typescript multiple generic cannot infered properly

Question

In vscode with typescript

multiple generic not infered properly when to use User-Defined Type Guards

const arraySample = [1, [3, 4], ["s"]];

const isFlat = <T, U>(array: (T | T[] | U | U[])[]): array is (T | U)[] => {
  return !array.some(Array.isArray);
};

if (isFlat(arraySample)) {
  arraySample;
}

vscode tooltip say below with error:

const isFlat: <number, number>(array: (number | number[])[]) => array is number[]

but I think it must be

const isFlat: <number, string>(array: (number | number[] | string | string[])[]) => array is (number|string)[]

could you let me know the proper solution?

jcalz · Accepted Answer · 2019-06-06 14:28:02Z

2

The compiler can't really infer anything meaningful for U in your code, since it won't know where to split up a union. Instead, since conditional types were introduced I'd tend to do it the other way... just let T be the element type of the array and pull out array types from it in the type guard. Like this:

const isFlat = <T>(array: T[]): array is Exclude<T, any[]>[] => {
    return !array.some(Array.isArray);
};

const arr = ["a", 2, true, Math.random() < 0.5 ? "d" : ["d"]];
if (isFlat(arr)) {
    arr; // (string | number | boolean)[]
} else {
    arr; // (string | number | boolean | string[])[]
}

Does that work? Hope that helps; good luck!

answered Jun 6, 2019 at 14:28

jcalz

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