Incorrect error on legitimate type overlap

TypeScript Version: 4.1.0-dev-20201102

Search Terms:

This condition will always return ‘false’ since the types ‘X’ and ‘Y’ have no overlap.

Code

function fn<T>(t: T) {
    return t === '' // error
}

fn('') // but here we pass it

https://www.typescriptlang.org/play?jsx=0#code/GYVwdgxgLglg9mABMMAeAKgPgBRQFyLoCUiA3gFCJWIBOAplCDUlIgLweIDkXiA9H0R0aNODXIBfcuRTYeJAYgDuYgNYBnIA

Expected behavior:

It works when we extend unknown. While this works on simple generics, it becomes inconvenient when the type parameter is nested deeper in a type structure.

function fn<T extends unknown>(t: T) {
    return t === '' // works
}

fn('') // works

https://www.typescriptlang.org/play?jsx=0#code/GYVwdgxgLglg9mABMMAeAKogpgDylsAEwGdFwBrMOAdzAD4AKKALkXQEpEBvAKEX8QAnLFBCCkURAF4ZiAORzEAeiWJqcQeWI8Avjx4oGCzirUatQA

Actual behavior:

This condition will always return 'false' since the types 'T' and 'string' have no overlap.

Related issues https://github.com/microsoft/TypeScript/issues/27910

[millsp]

Just of course the type information string is lost and t is T instead of T & string

That’s exactly what I’m complaining about.

The same would apply without the additional type-guard

I get that a generic T intersected with any or unknown yields T.

I’m pointing out the fact that we could safely narrow T with the type that sits on the rhs of the comparison operator.

But @RyanCavanaugh seems to say that it’s not possible, and I would like to know why should we have such a limitation.

Another example:

function fn<A, B>(a: A, b: B) {
    if (a === b) { // fails here
		// ...
    } else {	
		// ...
    }
}

const equals = <A>(thing: unknown, other: A): thing is A => {
    return thing === other;
};

function fn<A, B>(a: A, b: B) {
    if (equals(a, b)) { // works
        // a: A & B
    } else {
        // a: A
    }
}

[RyanCavanaugh]

A generic type parameter is comparable to the type of its constraint if a constraint was explicitly declared