Determining the return type based on an optional generic type in TypeScript

Question

Determining the return type based on an optional generic type in TypeScript

I have created a function that generates an object (map) [key] : value from an array.
My goal is to make the value method optional, and if not provided, simply return the item as it is.
Here is the code I have written:

   export default class ArrayUtils {
      /** Return a map from array like { [key]: value } */
      static toMap = <T, T2 = void, V = T2 extends void ? T : T2>(
        array: T[],
        key: (item: T) => string | number,
        value: (item: T) => V = (item: T) => item,
      ): { [key: string | number]: V } => Object.fromEntries(array.map((item) => [key(item), value(item)]));
    }

However, I am encountering an error in my IDE:

Type '(item: T) => T' is not assignable to type '(item: T) => V'.
Type 'T' is not assignable to type 'V'. 'V' could be instantiated with an arbitrary type which could be unrelated to 'T'

I have attempted to overload the function without success. How can I achieve the desired functionality?

typescript generics option-type

Answer 1

Answer №1

If you want to bypass the compiler's error, one way is to forcefully cast your T type to V using the method:

(item: T) => item as unknown as V

. However, a more appropriate solution would be to specify that your value method can return either V or T types like this: value: (item: T) => T | V

It's important to note that this error indicates a potentially unsafe conversion, as

'V' could be instantiated with an arbitrary type which may not be related to 'T'

.

It seems unnecessary to have 4 generic types for this method. Why not simply translate a T item to both a K and V, considering that the key will be of type string | number.

Here is the revised version:

static toMap<T, V>(
    array: T[],
    key: (item: T) => string | number,
    value: (item: T) => T | V = (item: T) => item,
): { [key: string | number]: V } {
  return Object.fromEntries(array.map((item) => [key(item), value(item)]));
}

Answer 2

If you want to bypass the compiler's error, one way is to forcefully cast your T type to V using the method:

(item: T) => item as unknown as V

. However, a more appropriate solution would be to specify that your value method can return either V or T types like this: value: (item: T) => T | V

It's important to note that this error indicates a potentially unsafe conversion, as

'V' could be instantiated with an arbitrary type which may not be related to 'T'

.

It seems unnecessary to have 4 generic types for this method. Why not simply translate a T item to both a K and V, considering that the key will be of type string | number.

Here is the revised version:

static toMap<T, V>(
    array: T[],
    key: (item: T) => string | number,
    value: (item: T) => T | V = (item: T) => item,
): { [key: string | number]: V } {
  return Object.fromEntries(array.map((item) => [key(item), value(item)]));
}

Answer 3

Answer №2

Update the value parameter to be optional by adding a ? after its name and change the return type of value from T to V | T.
If value is provided, then call value(item) to match the updated return type (V | T), if not, simply return item.

export default class ArrayUtils {
  /** Converts an array into a map like { [key]: value } */
  static toMap = <T, K extends string | number, T2 = void, V = T2 extends void ? T : T2>(
    array: T[],
    key: (item: T) => K,
    value?: (item: T) => V | T,
  ): { [key: string | number]: V | T } => Object.fromEntries(array.map((item) => [key(item), value ? value(item) : item]));
}

Answer 4

Update the value parameter to be optional by adding a ? after its name and change the return type of value from T to V | T.
If value is provided, then call value(item) to match the updated return type (V | T), if not, simply return item.

export default class ArrayUtils {
  /** Converts an array into a map like { [key]: value } */
  static toMap = <T, K extends string | number, T2 = void, V = T2 extends void ? T : T2>(
    array: T[],
    key: (item: T) => K,
    value?: (item: T) => V | T,
  ): { [key: string | number]: V | T } => Object.fromEntries(array.map((item) => [key(item), value ? value(item) : item]));
}

Answer 5

Answer №3

After some tweaking, I finally came up with this revised version of the code:

// type Key = string | number;
  /** Creates a map from an array in the format { [key]: value } */
  static toMap<T, V>(array: T[], key: (item: T) => Key, value: (item: T) => V): { [key in Key]: V };
  static toMap<T>(array: T[], key: (item: T) => Key): { [key in Key]: T };
  static toMap<T, V = void>(array: T[], key: (item: T) => Key, value?: (item: T) => V): { [key in Key]: V | T } {
    return Object.fromEntries(array.map((item) => [key(item), value ? value(item) : item]));
  }

Answer 6

After some tweaking, I finally came up with this revised version of the code:

// type Key = string | number;
  /** Creates a map from an array in the format { [key]: value } */
  static toMap<T, V>(array: T[], key: (item: T) => Key, value: (item: T) => V): { [key in Key]: V };
  static toMap<T>(array: T[], key: (item: T) => Key): { [key in Key]: T };
  static toMap<T, V = void>(array: T[], key: (item: T) => Key, value?: (item: T) => V): { [key in Key]: V | T } {
    return Object.fromEntries(array.map((item) => [key(item), value ? value(item) : item]));
  }

Determining the return type based on an optional generic type in TypeScript

Answer №1

Answer №2

Answer №3

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