Fastest method to invoke a potentially undefined function

With a background in C#, I am familiar with the null-conditional operator which allows you to call a function while avoiding a potential null-reference exception like this:

Func<int> someFunc = null;
int? someInteger = someFunc?.Invoke();
// someInteger == null

Considering that Typescript has its own "optional chaining operator" `.?` with similar functionality, I'm curious if there is a way to achieve the same concise code. The closest solution I can think of involves using a conditional expression:

let someFunc: (() => number) | undefined = undefined;
let someNumber = someFunc !== undefined ? someFunc() : undefined;

Maybe there is a way to leverage apply and call in this scenario?

Answer №1

Typescript allows for conditional invocation using optional chaining, a feature introduced in version 3.7.

const a = () => {console.log('hey')}
const b = null
a?.()
b?.()

Although the linter may raise some warnings, the code still complies and runs correctly. Check out this playground for a demonstration.

To learn more about optional chaining in Typescript, you can read about it in this blog post or in the official documentation.

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