TypeScript 4.1+: Generic binding too broad in recursive conditional types

TypeScript Version: 4.2.0-dev.20201103

Search Terms: Recursive Conditional types, generics, tuple types

Code

// A test object
const obj = {
    a: '2',
    b: {
        c: 3,
        d: {
            e: 'string',
            f: {
                g: {
                    h: 4
                }
            }
        }
    }
} as const

// its type
type Struct = typeof obj;

/**
 * First recursive conditional type 
 * 
 * CheckArguments gets 
 * Obj - A nested object
 * Arugments - A tuple of keys to go down a nested path
 * 
 * The type is recursive, I check if the current argument lists extends keyof Obj --> Then the recursion ends
 * Otherwise, I check if the first value in the tuple is keyof Obj, infer the rest, and go down the same type
 * again
 */
type CheckArguments<Obj, Arguments> = 
    Arguments extends [keyof Obj] ? Obj[Arguments[number]] :
    Arguments extends [keyof Obj, ...infer U] ? CheckArguments<Exclude<Obj[keyof Obj], string | number>, U> : never;

/**
 * Tests, all 👍
 */
type Foo = CheckArguments<Struct, ['a']> // "2"
type Foo2 = CheckArguments<Struct, ['b', 'd', 'e']> // "string"
type Foo3 = CheckArguments<Struct, ['b', 'd', 'f', 'g', 'h']> // 4

/**
 * Second recursive conditional type
 * 
 * Arguments gets
 * Obj - A nested Obj
 * 
 * The recursive conditional type creates a union type of possible nested arguments in a tuple
 * This is based on Anders example from TSConf: https://github.com/ahejlsberg/tsconf2020-demos/blob/master/template/main.ts
 * (Dotted paths)
 */
type Arguments<Obj> = 
    Obj extends object ?
        [keyof Obj] | SubArguments<Obj, keyof Obj> :
        never;

// A helper type
type SubArguments<Obj, Key> =  Key extends keyof Obj ? [Key, ...Arguments<Obj[Key]>] : never;

// For example, the possible tuples of Struct 👍
type Bar = Arguments<Struct>;

// equals to this union type
type Bar2 = ["a" | "b"] | ["b", "c" | "d"] | ["b", "d", "e" | "f"] | ["b", "d", "f", "g"] | ["b", "d", "f", "g", "h"]

/**
 * So both recursive conditional types work on their own. A problem is once I want to combine
 * them in a function, where I expect the first argument to bind to a value type within the Arguments union
 * 
 * Instead of having just one value type passed to CheckArguments (the one that is bound through the generic Keys),
 * TypeScript passes all parts of the union to CheckArguments. This leads CheckArguments to return all possible values
 */

declare function get<Obj extends object, Keys extends Arguments<Obj>>(o: Obj, ...keys: Keys): CheckArguments<Obj, Keys>

/** 
 * Tests 💥
 * */
const foo = get(obj, 'a') // Should be "2" 😢 
const foo1 = get(obj, 'b', 'c') // Should be  3 😢 
const foo2 = get(obj, 'b', 'd', 'e') // Should be "string" 😢 
const foo3 = get(obj, 'b', 'd', 'f', 'g', 'h') // Should be 4 😢 

Expected behavior: Keys gets bound to the value type passed as an argument to the function. This value type is then used for CheckArguments

Actual behavior: Keys is the entire union type Arguments<Obj>, not the subset. This leads to CheckArguments returning a too broad return type (and taking very long to evaluate ;-))

Playground Link: Click here

Related Issues: Did not find any.

Beginne damit, den generischen Inferenzfall im verknüpften TypeScript Playground mit der gemeldeten Version 4.2.0-dev.20201103 zu reproduzieren, und vergleiche die inferierten Typen von foo bis foo3 mit den erwarteten Ergebnissen. Als erledigt gilt die Aufgabe, wenn jeder Aufruf Keys an das bereitgestellte Tupel statt an die vollständige Arguments-Union bindet und dabei die beschriebene übermäßige Auswertung vermeidet.