Why can't I use an anonymous function expression for accessing an array index position?

user_12834972 0
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In JavaScript one can access array values using this syntax...

const arr = ["one", "two", "three"];
console.log(arr[0]); // "one"

...or using a defined function that returns a number to achieve the same result:

const arr = ["one", "two", "three"];
const fn = () => 0;
console.log(arr[fn()]); // "one"

I might be brain-afk, but I simply do not understand why the exact same thing does not work when using an anonymous function expression? Don't I clearly end up with 0 as an array index position assessor?

const arr = ["one", "two", "three"];
console.log(arr[() => 0]); // undefined
anonymous-function function arrays javascript
2 Answers

() => 0 returns a function, not the value returned when called.

You need to invoke the function instead:

const arr = ["one", "two", "three"];
console.log(arr[(() => 0)()]);

user_142512210

The main difference is about the concept of an anonymus function.

const arr = ["one", "two", "three"];
const fn = () => 0;
console.log(arr[fn()]);

Here you are assigning an anonymus function to the fn constant and you are invoking this function returning a 0 in the index.

In your last example you're just assigning the function without invoking the function so that's why it returns the undefined.

If you want to auto invoke an anonymous function you just need to call it like this:

(() =>0)() // check the last pair of parentheses
user_137152680
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