Proposal: better type narrowing in overloaded functions through overload set pruning

TypeScript Version: 2.7.2

Search Terms: type inference, type narrowing, overloads

Code

function bar(selector: "A", val: number): number;
function bar(selector: "B", val: string): string;
function bar(selector: "A" | "B", val: number | string): number | string
{
    if (selector === "A")
    {
        const the_selector = selector;  // "A"
        const the_val = val;            // number | string but should be just number
        return "foo";                   // should give error: when selector is "A" should return a number
    }
    else
    {
        const the_selector = selector;  // "B"
        const the_val = val;            // number | string but should be just string
        return 42;                      // should give error: when selector is "B" should return a string
    }
}

Issue

The snippet above shows a limitation of the current type inference implementation in TypeScript.
As can be clearly seen from the overloaded declarations, when selector === "A" we have val: number and we must return a number, while when selector === "B" we have val: string and we must return a string. But the type-checker, although able to understand that the_selector can only be "A" in the if block and only "B" in the else block, still types the_val as number | string in both blocks and allows us to return a string when selector === "A" and a number when selector === "B".

Possible Solution

In the implementation of an overloaded function, when the type-checker decides to narrow the type of a parameter (as often happens in if-else blocks), it should also remove the overloads that don't match the narrowed parameter type from the overload set, and use the pruned overload set to narrow the types of the other parameters.

Playground Link: https://www.typescriptlang.org/play/#src=%0Afunction%20bar(selector%3A%20%22A%22%2C%20val%3A%20number)%3A%20number%3B%0Afunction%20bar(selector%3A%20%22B%22%2C%20val%3A%20string)%3A%20string%3B%0Afunction%20bar(selector%3A%20%22A%22%20%7C%20%22B%22%2C%20val%3A%20number%20%7C%20string)%3A%20number%20%7C%20string%0A%7B%0A%20%20%20%20if%20(selector%20%3D%3D%3D%20%22A%22)%0A%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20const%20the_selector%20%3D%20selector%3B%0A%20%20%20%20%20%20%20%20const%20the_val%20%3D%20val%3B%0A%20%20%20%20%20%20%20%20return%20%22foo%22%3B%0A%20%20%20%20%7D%0A%20%20%20%20else%0A%20%20%20%20%7B%0A%20%20%20%20%20%20%20%20const%20the_selector%20%3D%20selector%3B%0A%20%20%20%20%20%20%20%20const%20the_val%20%3D%20val%3B%0A%20%20%20%20%20%20%20%20return%2042%3B%0A%20%20%20%20%7D%0A%7D

Commencez par reproduire l’exemple de fonction surchargée dans le TypeScript Playground lié et examinez le comportement du vérificateur de types pour les paramètres restreints. Comparez les types inférés et les instructions return acceptées dans chaque branche. C’est terminé lorsque l’ensemble des overloads est restreint de manière cohérente, afin que le paramètre associé et le type de retour soient vérifiés par rapport aux overloads restants.