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

Beginne damit, das Beispiel für eine überladene Funktion im verlinkten TypeScript Playground zu reproduzieren, und untersuche das Verhalten des Type-Checkers bei eingegrenzten Parametern. Vergleiche die abgeleiteten Typen und die akzeptierten return-Anweisungen in jedem Zweig. Als erledigt gilt die Aufgabe, wenn die Menge der Overloads konsistent eingegrenzt wird, sodass der zugehörige Parameter und Rückgabetyp anhand der verbleibenden Overloads geprüft werden.