Javascript: How to pass an array with specific index to a function?

Question

Let say I have an array:-

var arr = [0,1,2,3,4]

I need to pass only "1" and "4" to function below

function func1(onlyArray){
 //Do Stuff...
}

I have tried both but both also dont work

func1(arr[1,4])
func1(arr[[1],[4]])

Can anyone show me the right way or give me some keyword?

• Egor Stambakio: func(arr[1], arr[4])
• vbnewbie: @wostex but the function only accept one parameter, will this detect as two parameter?
• Mihai Alexandru-Ionut: @vbnewbie, use this: func([arr[1], arr[4]])
• davidbuzatto: What is the type of the parameter?
• j08691: Do you have control over the function or are you locked into it only accepting a single argument? If you can't change the function, how will the function know how many parameters it's receiving?

Answer 1

You can use this: func([arr[1], arr[4]])

We are taking the elements at index 1 and 4 of the array arr and creating a new array with those elements. Then we pass that newly created array to func.

Answer 2

If you need a single array instead of 2, use this:

var arr = [0,1,2,3,4];
function func1(onlyArray){
 //Do Stuff...
 console.log(onlyArray); // [1, 4]
}
func1(arr.filter((item,index) => index === 1 || index === 4));

• doubleOrt: [arr[1], arr[4]] is much simpler, why not just use that ?
• Mihai Alexandru-Ionut: @Taurus, he just provided another solution.
• doubleOrt: @Alexandru-IonutMihai Oh
• colecmc: @Taurus, I've included an optional way to get a single array. Yours works as well.
• Mouser: I think this better fits the op's question. Since you want to pass a single array consisting only of 1 and 4.

Answer 3

So using your array:

var arr = [0,1,2,3,4];

and your function:

function myFunction(newArr){ //Do stuff console.log(newArr); };

You can wrap the array indexes you want to use inside of their own array when you call the function, like the following:

myFunction([arr[1], arr[4]]); //Output is [1, 4]

• chazsolo: How is this different than @Taurus's answer?
• Trevor W.: It isn't. Their answer is more brief and earlier than my comment. This is my first time trying to help on StackOverflow, so I spent more time learning how to use the markup than I really did responding. Sorry, my friend.
• chazsolo: No problem - thanks for the explanation, and your willingness to help!