Consider the following use of extends to remove members from a union type:

type RemoveNames<N, K> = N extends K ? never : N

type Names = 'bill' | 'jane' | 'freddy'    
type Bill = RemoveNames<Names, 'jane' | 'freddy'> // type Bill = 'bill'

In a related question, N extends K was defined to mean that "N is assignable to K", but in this case (on the face of it) N is not assignable to K - only a narrowed version of N is assignable to K.

How to interpret extends in this case? Are the members of N iterated over and "filtered" (?) if they're not in K?

CodePudding user response：

When resolving the Bill type, TypeScript takes every type of the Names union type and resolves the conditional type N extends K ? never : N, then builds a final union type with the results:

• Does 'bill' extend 'jane' | 'freddy'? No: N extends K ? never : N gives 'bill'
• Does 'jane' extend 'jane' | 'freddy'? Yes: N extends K ? never : N gives never
• Does 'freddy' extend 'jane' | 'freddy'? Yes: N extends K ? never : N gives never

The final type is the union of these three types 'bill' | never | never which gives 'bill'.

This behavior is described in the Distributive Conditional Types chapter of the docs.