Is it possible to include a conditional type in TypeScript using inlining?

Apologies if this question seems basic to those well-versed in typesystems. I'm puzzled by the difference in outcomes when inlining a conditional statement in TypeScript.

For instance, using the Extract conditional leads to different results depending on whether it's inlined or used as a generic:

// Initial setup
type TUnion = ['foo', 1] | ['baz', 2];
type TDiscriminator = ['foo', any];

// Using conditional generic
type Extract<T, U> = T extends U ? T : never;
type TExtracted = Extract<TUnion, TDiscriminator>; // ["foo", 1]

// Inlining the conditional
type TInlined = TUnion extends TDiscriminator ? TUnion : never; // never

Upon further exploration, it appears that passing the value of T is necessary:

type ExtractWithU<T> = T extends TDiscriminator ? T : never;
type TExtractedWithU = ExtractWithU<TUnion>; // ["foo", 1]

type ExtractWithT<U> = TUnion extends U ? TUnion : never;
type TExtractedWithT = ExtractWithT<TDiscriminator>; // never

Could I be overlooking something quite simple here?

Answer №1

After reviewing the conversation between jcalz and Titian above, it became evident that the issue lies in the distribution of the generic versus the inlined version.

To illustrate this point:

// setting up
type TA = ['foo', 1];
type TB = ['baz', 2];
type TUnion = TA | TB;
type TDiscriminator = ['foo', any];

// using the conditional generic
type Extract<T, U> = T extends U ? T : never; 
type TExtracted = Extract<TUnion, TDiscriminator>; // ["foo", 1]
//              = Extract<TA | TB, TDiscriminator>
//              = Extract<TA, TDiscriminator> | Extract<TB, TDiscriminator>
//              = TA | never
//              = TA

// inlining the conditional
type TInlined = TUnion extends TDiscriminator ? TUnion : never; // never
//            = never, as the union does not match the discriminator

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