Function parameter type inference error about typescript HOT 4 CLOSED

Duplicate #33014

For now, if you need this behavior you either need to rewrite in terms of #47109 and use generic indexing instead of control flow anlaysis:

function func<T extends Type>(type: T, extend: X<T>) {
    const x: { [K in Type]: (type: K, x: X<K>) => void } = {
        [Type.A](type, extend) {
            console.log(type, extend);
            //                ^? (parameter) extend: string
        },
        [Type.B](type, extend) {

        }
    }
    x[type](type, extend);
}
func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // still allowed, #47109 technically unsound this way

or, you need to use control flow analysis on a discriminated union and not use generics:

function func(...[type, extend]: { [T in Type]: [type: T, extend: X<T>] }[Type]) {
    if (type === Type.A) {
        console.log(type, extend);
        //                ^? (parameter) extend: string
    }
}

func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // error

Playground link to code

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Because checking what value type is does not have any relevance for what type the value of extend is.

This is a perfectly valid call: func<Type.A | Type.B>(Type.A, 1);

from typescript.

Because checking what value type is does not have any relevance for what type the value of extend is.

This is a perfectly valid call: func<Type.A | Type.B>(Type.A, 1);

How should I constrain the type of "extend" ? Do you have any suggestions ? Thank you so much !

from typescript.

Duplicate #33014

For now, if you need this behavior you either need to rewrite in terms of #47109 and use generic indexing instead of control flow anlaysis:

function func<T extends Type>(type: T, extend: X<T>) {
    const x: { [K in Type]: (type: K, x: X<K>) => void } = {
        [Type.A](type, extend) {
            console.log(type, extend);
            //                ^? (parameter) extend: string
        },
        [Type.B](type, extend) {

        }
    }
    x[type](type, extend);
}
func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // still allowed, #47109 technically unsound this way

or, you need to use control flow analysis on a discriminated union and not use generics:

function func(...[type, extend]: { [T in Type]: [type: T, extend: X<T>] }[Type]) {
    if (type === Type.A) {
        console.log(type, extend);
        //                ^? (parameter) extend: string
    }
}

func(Type.A, '');
func(Type.B, 1);
func(Math.random() < 0.99 ? Type.A : Type.B, 1); // error

Playground link to code

Good job

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