How can I only accept one of two specific function signatures in Typescript and reject any others?

Looking for an answer from the community on Typescript, require either of two function signatures

In my code, I am handling a callback where I need to either pass an error or leave the first argument as undefined and pass data as the second argument.

To describe this, I have created an interface as shown below:

type Callback = {
  (error: undefined, value: string): void;
  (error: Error): void;
}

function doThings(c: Callback) {
    // Valid scenario with no error and a useful value.
    c(undefined, '1');
    // Scenario where something went wrong, so we provide an error but no value.
    c(new Error());

    // TS playground accepts this which is not ideal:
    c(undefined);
}

function cOverload(error: undefined, value: string): void;
function cOverload(error: Error): void;
function cOverload(error: undefined | Error, value?: string) { }

doThings(cOverload)

I am facing two issues in this code, both related to how I define the function signature(s).

  1. c(undefined); is not flagged as an error by TS, but I want it to be treated as one.
  2. In the implementation of cOverload, the parameter value is considered a string. However, I do not need to check its type, and TS allows me to use functions like { value.charAt(0); }, while I expect it to raise concerns about value potentially being undefined.

Answer №1

  1. The compiler allows the 3rd invocation c(undefined) because TypeScript types are nullable by default. Enabling the strictNullChecks compiler setting (found in the Options dropdown of the Playground) will reveal that this invocation results in an error.

  2. Similar to the previous answer, enabling strictNullChecks is also beneficial here.

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