Type aliases using generics may demonstrate varying behaviors from type aliases without generics

Here is a code snippet to consider:

type TestTuple = [
    { test: "foo" },
    {
        test: "bar";
        other: 1;
    }
];

type Foo<Prop extends string> = TestTuple extends Record<Prop, string>[]
    ? true
    : false;
type X = Foo<"test">;

type Prop = "test";
type Y = TestTuple extends Record<Prop, string>[]
    ? true
    : false;

// X is type false
const x: X = false;
// Y is type true
const y: Y = true;

Playground link.

Types Foo and Y are quite similar. However, the main difference is that Foo has a generic parameter Prop, while Y simply uses a type alias named Prop. Even though it's not necessary, I wanted their declarations to mirror each other exactly. Therefore, shouldn't Foo<"test"> (represented by the type X) and Y have the same outcomes? Surprisingly, they do not. X results in type false, whereas Y evaluates to type true. Interestingly, altering the other property in

TestTuple</code to a string or removing it completely causes both <code>X
and Y to evaluate as true, which aligns with expectations.

My question then becomes: why is this discrepancy happening? Could it be a bug in the compiler? If so, has an issue already been reported but remains elusive to me? Or, perhaps this behavior is related to how generics are dealt with in TypeScript?

Answer №1

UPDATE: Good news - TypeScript 4.2 has resolved this issue! Check out the Playground link to updated code


I've reported microsoft/TypeScript#41613 regarding this issue by simplifying it to the following example:

type What<K extends string> =
    { x: { y: 0, z: 1 } } extends { x: { [P in K]: 0 } } ? true : false;

type Huh = What<"y">; // expecting true but receiving false!

The lead architect for TypeScript, Anders Hejlsberg, shared his insights in a comment:

In determining whether to defer resolution of the conditional type, we evaluate the "most permissive instantiations" of the check and extends types. The constraint of the most permissive instantiation of the extends type is currently { x: { [index: string]: 0 } }, when it should be { x: { } }. This is an easy fix that will be included in the upcoming PR.

Hopefully, this fix will be implemented in an upcoming PR and incorporated into TypeScript 4.2 (Update: it has been merged). Once integrated, this should address the issue you raised, ensuring that instead of enclosing the indexed type with {x: ...}, it will be wrapped with a tuple type.

For now, you may want to consider utilizing a workaround like the one suggested in @Temoncher's response.

Playground link to updated code

Answer №2

Regrettably, I am not familiar with this issue personally. It seems to be related to dictionary types such as Record or { [K in Prop]: any }, as it only fails for these types based on my own observations. Hopefully, someone will be able to provide a more comprehensive solution.

However, I can suggest a workaround.

Solution

Instead of directly comparing TestTuple with Record<...>[], consider comparing TestTuple[number] with Record<...> instead

type Bar<P extends string> = TestTuple[number] extends Record<P, string>
  ? true
  : false
// returns true
type A = Bar<'test'>

// also results in true
type B = TestTuple extends Record<'test', string>[]
    ? true
    : false;

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