Strategies for handling TypeScript's treatment of uninitialized optional parameters as non-undefined values

Here is a simplified version of my function:

const nullify = <T extends string | undefined>(
  value?: T,
): T extends string ? string : null => {
  if (value === undefined) return null as any;
  return value as any;
};

I expected this function to handle lack of value as undefined, but TypeScript does not allow it and returns a union of string or null. To retrieve null from this function, you must specifically pass undefined.

const result1 = nullify('string'); // result1 is a string, which is the expected behavior
const result2 = nullify(undefined); // result2 is null, as expected
const result3 = nullify(); // result3 is a combination of string and null, but ideally should be just null

Is there a standard way to address this issue?

Answer №1

One issue arises when you omit the value parameter; the compiler struggles to infer type T and defaults to the constraint string or undefined. To alter this default behavior, set a default value for the generic parameter using the = syntax:

const nullify = <T extends string | undefined = undefined>( 
//                      default generic ----> ^^^^^^^^^^^
    value?: T,
): T extends string ? string : null => {
    if (value === undefined) return null as any;
    return value as any;
};

The difference becomes noticeable only when value is omitted:

const result1 = nullify('string');  // returns 'string' as expected
const result2 = nullify(undefined); // returns null as expected
const result3 = nullify();  // returns null as expected 👍

Great job!


The drawback is that technically, the caller can specify any value for T they desire. So, an eccentric caller could do this:

const oops = nullify<string>(); // returns 'string', oops
// ---------------> ^^^^^^^^ <--- manual specification of string

Since string is manually specified for T, and the value parameter is optional, there are no compiler errors. As a result, the compiler mistakenly considers a null value as being of type string.

Although this scenario is unlikely, you may choose to ignore it. TypeScript is actually unsound in several areas, so you wouldn't be worsening things significantly here.

However, if accuracy matters to you, consider refactoring. One approach is to align your generic type parameter with the tuple type of a rest parameter:

const nullify = <T extends [string] | [undefined?]>(
    ...[value]: T
): T extends [string] ? string : null => {
    if (value === undefined) return null as any;
    return value as any;
};

const result1 = nullify('string');  // returns 'string' as expected
const result2 = nullify(undefined); // returns null as expected
const result3 = nullify();  // returns null as expected

In this approach, passing an incorrect value becomes less straightforward since only cases where value is missing or undefined and T expands [undefined?] are acceptable:

const okayNow = nullify<[string]>(); // error thrown, as expected

Alternatively, instead of generics, you can utilize overloads to achieve similar results. For more information on setting default values for generic parameters, check out How to assign a default value to a generic parameter?.

Access the code in the Playground

Answer №2

Maybe consider verifying if the data type is a string.

const nullify = <T extends string | undefined>(
  value?: T,
): T extends string ? string : null => {
  if (typeof value === 'string') return value as any;
  return null as any;
};

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