Why do JavaScript functions work at different times?

Question

Bit of a weird title, let me explain.

Take this:

let a = ['a', 'b', 'c']
function joining(){
    console.log(a.join())
}
joining() // a,b,c
THE ABOVE IS JOINED

function joining(){
    a.join()
    console.log(a)
}
joining() // ['a', 'b', 'c']
THE ABOVE IS NOT JOINED

function pushing(){
    a.push('d')
    console.log(a)
}
pushing() // ['a', 'b', 'c', 'd']
THE ABOVE IS PUSHED

The first joining() joins them correctly as I have called .join() on the array. However, the second one I have called joined on it and then tried to log it but it displays the original. However I did exactly the same with the push() method and that changed the original array. how can I tell what methods will change the original array and which wont or is it just a case of learning each method?

@Pyromonk that's my question. do i need to learn each individual method to see if it alters the original or not? — The worm2
– The worm2, Commented Jun 1, 2017 at 12:16
@Theworm2 Only arrays and objects can be mutated. So any function that returns some other type can't possibly alter the original to do it. — Barmar
– Barmar, Commented Jun 1, 2017 at 12:18

Scott · Accepted Answer · 2017-06-01 12:18:58Z

2

The reason that a.join() does not output your expected result in the second instance is because you have not assigned the returned value from the invocation of the method.

a = a.join();
console.log(a);

For completeness, in your first example the expected output is logged on the console because the returned value from the invocation is passed as a parameter to console.log().

answered Jun 1, 2017 at 12:18

Scott

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omri_saadon · Accepted Answer · 2017-06-01 12:16:04Z

1

The join() method joins the elements of an array into a string, and returns the string.

Therefore, you should update it into:

a = a.join()

answered Jun 1, 2017 at 12:16

omri_saadon

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