Implying generics at a later time, not during instantiation / altering the type of a generic

Question

Implying generics at a later time, not during instantiation / altering the type of a generic

If we consider the example provided, is there a way to instruct the typescript compiler that the return type of baz must be string, since it can be inferred from foo.a('aString') that it's a string?

const fn = <T,S>()=>{
  let s: S
  let t: T
  return {
    a: <Z extends T=T>(arg: Z)=>{
      t = arg
      return s
    },
    b: <V extends S=S>(arg:V)=>{
      s = arg
      return t
    }
  }
}
const foo = fn()
const bar = foo.a('aString')
// currently,baz's type is 'unknown', but it is determinable that the type should be 'string'
const baz = foo.b('aString')

code

A solution is to define

const foo = fn<string, string>()

. However, this requires one to specify both types upfront when foo.a('aString') may be the best place to infer T, rather than having to explicitly specify the type.

I suspect not ... but one can have hopium!

typescript

Answer 1

Answer №1

fn functions simulate the behavior of a class. TypeScript provides strong support for classes, so it is recommended to utilize class in your code examples. Take a look at this implementation:

class fn<T, S>{
  s: S | undefined = undefined
  t: T | undefined = undefined

  constructor() { }

  a<Z extends T = T>(arg: Z): asserts this is this & { t: Z } {
    this.t = arg
  }
  b<V extends S = S>(arg: V): asserts this is this & { s: V } {
    this.s = arg
  }

}
const foo: fn<string, string> = new fn<string, string>()

foo.a('aString')
foo.t // 'aString'

foo.b('bString')
foo.s // 'bString'

Interactive Playground

If you modify this within your class and need to track those changes, you must use assertion functions.

However, this approach won't work if you call the fn function without explicitly specifying generic arguments. In such cases, TypeScript defaults to using unknown as the type.

Consider the following example:

const bar = <T,>() => {
  return 42 as any as T
}
//const bar: <unknown>() => unknown
bar()

The T type cannot be inferred because no value was provided. The same issue occurs with the fn function as well.

Answer 2

fn functions simulate the behavior of a class. TypeScript provides strong support for classes, so it is recommended to utilize class in your code examples. Take a look at this implementation:

class fn<T, S>{
  s: S | undefined = undefined
  t: T | undefined = undefined

  constructor() { }

  a<Z extends T = T>(arg: Z): asserts this is this & { t: Z } {
    this.t = arg
  }
  b<V extends S = S>(arg: V): asserts this is this & { s: V } {
    this.s = arg
  }

}
const foo: fn<string, string> = new fn<string, string>()

foo.a('aString')
foo.t // 'aString'

foo.b('bString')
foo.s // 'bString'

Interactive Playground

If you modify this within your class and need to track those changes, you must use assertion functions.

However, this approach won't work if you call the fn function without explicitly specifying generic arguments. In such cases, TypeScript defaults to using unknown as the type.

Consider the following example:

const bar = <T,>() => {
  return 42 as any as T
}
//const bar: <unknown>() => unknown
bar()

The T type cannot be inferred because no value was provided. The same issue occurs with the fn function as well.

Answer 3

Answer №2

Here is a clever workaround:

const func = <T,S>(t?: T, s?: S)=>{
  return {
    a<NewT>(arg: NewT) {
      return func<NewT,S>(arg,s)
    },
    b<NewS>(arg: NewS) {
      return func<T,NewS>(t,arg)
    },
    t(){
      return t as T // changed for simplicity
    },
    s(){
      return s as S // changed for simplicity
    }
  }
}
const example1 = func()
const example2 = example1.a('aString')
const result1 = example2.t() // string
const result2 = example2.s() // unknown
const example3 = example1.b('aString') 
const result3 = example3.t() // unknown
const result4 = example3.s() // string
const finalExample = example2.b('aString') 
const result5 = finalExample.t() // string
const result6 = finalExample.s() // string

Answer 4

Here is a clever workaround:

const func = <T,S>(t?: T, s?: S)=>{
  return {
    a<NewT>(arg: NewT) {
      return func<NewT,S>(arg,s)
    },
    b<NewS>(arg: NewS) {
      return func<T,NewS>(t,arg)
    },
    t(){
      return t as T // changed for simplicity
    },
    s(){
      return s as S // changed for simplicity
    }
  }
}
const example1 = func()
const example2 = example1.a('aString')
const result1 = example2.t() // string
const result2 = example2.s() // unknown
const example3 = example1.b('aString') 
const result3 = example3.t() // unknown
const result4 = example3.s() // string
const finalExample = example2.b('aString') 
const result5 = finalExample.t() // string
const result6 = finalExample.s() // string

Implying generics at a later time, not during instantiation / altering the type of a generic

Answer №1

Answer №2

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