Can you provide an overview of the ternary operator within advanced method signatures in TypeScript?

Question

Can you provide an overview of the ternary operator within advanced method signatures in TypeScript?

I'm struggling to understand the method signature provided below:

export type UnionToIntersection<U> = (U extends any ? (k: U) => void : never) extends (k: infer I) => void ? I : never;

This code snippet was in response to a question on converting a union to an intersection type in TypeScript, which you can find here:

But I find the method signature quite confusing.

Let's try to break it down:

Understanding the Ternary Operator

(U extends any ? (k: U) => void : never)

Does this mean that if an object is provided, the expected argument should be a function (k: U) => void?

Using the 'extends' Keyword

(U extends any ? (k: U) => void : never) extends (k: infer I)

My assumption is that infer is used to deduce the type of I, but the rest is still unclear to me.

Any assistance in explaining this would be greatly appreciated. Here is a related discussion thread that I found confusing as well: https://github.com/Microsoft/TypeScript/issues/27907

typescript

Answer 1

Answer №1

Great job on nailing the first part, let's refer to that as T1:

export type T1<U> = (U extends any ? (k: U) => void : never);

Now, moving on to what remains (using parentheses for clarity):

export type UnionToIntersection<U> = (T1<U> extends (k: infer I) => void) ? I : never;

This segment once again utilizes the ternary operator with the condition

T1<U> extends (k: infer I) => void

. By incorporating the infer keyword, typescript can infer the type at this point and utilize it later in the type definition. Essentially, this part conveys:

"If T1 extends a function (k) => void, use the type of its parameter k, or else use never."

Therefore, the entire statement (assuming U is not

never</code) functions by creating a type <code>(k: U) => void

from U, solely to then infer the type of parameter

k</code, which seems to be inferred as a very specific type, resulting in the intersection of what <code>U

represents as a union of types.

Answer 2

Great job on nailing the first part, let's refer to that as T1:

export type T1<U> = (U extends any ? (k: U) => void : never);

Now, moving on to what remains (using parentheses for clarity):

export type UnionToIntersection<U> = (T1<U> extends (k: infer I) => void) ? I : never;

This segment once again utilizes the ternary operator with the condition

T1<U> extends (k: infer I) => void

. By incorporating the infer keyword, typescript can infer the type at this point and utilize it later in the type definition. Essentially, this part conveys:

"If T1 extends a function (k) => void, use the type of its parameter k, or else use never."

Therefore, the entire statement (assuming U is not

never</code) functions by creating a type <code>(k: U) => void

from U, solely to then infer the type of parameter

k</code, which seems to be inferred as a very specific type, resulting in the intersection of what <code>U

represents as a union of types.

Can you provide an overview of the ternary operator within advanced method signatures in TypeScript?

Understanding the Ternary Operator

Using the 'extends' Keyword

Answer №1

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