A TypeScript function with nested options and a variable rest parameter: The type 'T' is open to instantiation with various subtypes

Question

A TypeScript function with nested options and a variable rest parameter: The type 'T' is open to instantiation with various subtypes

Let's consider the following simplified example that showcases my issue:

const customFactory = <T extends unknown[]>(
  customFn: (...args: T) => void
) => {
  const modifiedCustomFn = (...args: T) => {
    console.log('Called with arguments:', ...args)
    return customFn(...args)
  }

  // initialization
  modifiedCustomFn() // TypeScript error occurs here

  return modifiedCustomFn
}

// implementation:
const customFunction = (numberValue?: number, stringValue?: string) => {
  return JSON.stringify({ numberValue, stringValue })
}
customFactory(customFunction)()

Upon execution, I encounter a TypeScript error message:

The argument type '[]' cannot be assigned to the parameter type 'T'. While '[]' fits within the constraint of type 'T', it's plausible for 'T' to potentially instantiate a distinct subtype other than 'unknown[]'.

If I were to provide this function instead:

const customFunction = <T extends unknown[]>(...args: T) => {
  return args
}
customFunction()

This is perplexing. How do I go about persuading TypeScript that both cases refer to the same T?

typescript

Answer 1

Answer №1

When you tell Typescript that the argument function possibly takes arguments but then call it with none, that's when it starts complaining. Imagine if you pass in a function that actually requires arguments - you'd end up with a runtime error, which defeats the purpose of static analysis.

To solve this issue, simply include an empty array as a type for the arguments to specify the type of the argument function:

const factory = <T extends unknown[], R>(
  fn: (...args: T | []) => R // Note []
) => {
  const transformedFn = (...args: T | []) => { // Note []
    console.log('called with', ...args)
    return fn(...args)
  }

  // initialization
  transformedFn() // No more errors

  return transformedFn
}

If you attempt to call factory with a function that has non-optional arguments, a type error will arise:

function foo(a: number): string {
    return a.toFixed(2)
}

factory(foo)() // error!

However, your initial call works without any issues:

const fn = (num?: number, str?: string) => {
  return JSON.stringify({ num, str })
}
factory(fn)()

You may also notice the addition of another generic R for the return type of the argument function. This is because although you specified the return type as void in your signature, in reality, you are returning a string value in your example. If you prefer not to return anything, you can revert it back to void and omit the second generic.

Playground

Answer 2

When you tell Typescript that the argument function possibly takes arguments but then call it with none, that's when it starts complaining. Imagine if you pass in a function that actually requires arguments - you'd end up with a runtime error, which defeats the purpose of static analysis.

To solve this issue, simply include an empty array as a type for the arguments to specify the type of the argument function:

const factory = <T extends unknown[], R>(
  fn: (...args: T | []) => R // Note []
) => {
  const transformedFn = (...args: T | []) => { // Note []
    console.log('called with', ...args)
    return fn(...args)
  }

  // initialization
  transformedFn() // No more errors

  return transformedFn
}

If you attempt to call factory with a function that has non-optional arguments, a type error will arise:

function foo(a: number): string {
    return a.toFixed(2)
}

factory(foo)() // error!

However, your initial call works without any issues:

const fn = (num?: number, str?: string) => {
  return JSON.stringify({ num, str })
}
factory(fn)()

You may also notice the addition of another generic R for the return type of the argument function. This is because although you specified the return type as void in your signature, in reality, you are returning a string value in your example. If you prefer not to return anything, you can revert it back to void and omit the second generic.

Playground

Answer 3

Answer №2

This code snippet has been crafted to address a specific issue:

const factory = <T extends unknown>(
  fn: (...args: T[]) => void
) => {
  const transformedFn = (...args: T[]) => {
    console.log('called with', ...args)
    return fn(...args)
  }

  // initialization
  transformedFn()

  return transformedFn
}

The T extends unknown constraint serves the purpose of preventing TypeScript Playground from mistaking it for React code.

The original implementation encountered problems because it allowed types that extended unknown[], but did not permit an empty array, as shown below:

function example<T extends unknown[]>(...args: T) {
}

const args = [1, 2, 3] as const;

example(args);

This situation rendered the call to transformedFn within the factory function invalid since there existed a type that met T extends unknown[] but failed to meet [] (the type expected when calling transformedFn without arguments).

An alternative approach could involve retaining the existing code structure while also passing in the necessary arguments for the transformedFn call - thereby upholding the original types:

const factory = <T extends unknown[]>(
  fn: (...args: T) => void,
  args: T
) => {
  const transformedFn = (...args: T) => {
    console.log('called with', ...args)
    return fn(...args)
  }

  // initialization
  transformedFn(...args)

  return transformedFn
}

Answer 4

This code snippet has been crafted to address a specific issue:

const factory = <T extends unknown>(
  fn: (...args: T[]) => void
) => {
  const transformedFn = (...args: T[]) => {
    console.log('called with', ...args)
    return fn(...args)
  }

  // initialization
  transformedFn()

  return transformedFn
}

The T extends unknown constraint serves the purpose of preventing TypeScript Playground from mistaking it for React code.

The original implementation encountered problems because it allowed types that extended unknown[], but did not permit an empty array, as shown below:

function example<T extends unknown[]>(...args: T) {
}

const args = [1, 2, 3] as const;

example(args);

This situation rendered the call to transformedFn within the factory function invalid since there existed a type that met T extends unknown[] but failed to meet [] (the type expected when calling transformedFn without arguments).

An alternative approach could involve retaining the existing code structure while also passing in the necessary arguments for the transformedFn call - thereby upholding the original types:

const factory = <T extends unknown[]>(
  fn: (...args: T) => void,
  args: T
) => {
  const transformedFn = (...args: T) => {
    console.log('called with', ...args)
    return fn(...args)
  }

  // initialization
  transformedFn(...args)

  return transformedFn
}

A TypeScript function with nested options and a variable rest parameter: The type 'T' is open to instantiation with various subtypes

Answer №1

Answer №2

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