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Home/ Questions/Q 9051469
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T12:54:44+00:00 2026-06-16T12:54:44+00:00

I have an array that contains variables and functions. The array is 80 elements

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I have an array that contains variables and functions. The array is 80 elements long. The first 20 elements are used together in a for loop. When the loop has completed, the first twenty elements are moved to the back of the array, and the for loop starts again.

I am rebuilding the array this way:

var a2=[the array with 80 elements];
run(a2);
function run(array){
  var n=array.slice(0,20); array.splice(0,20);
  var con=array.concat(n); a2=con;
  }

So I am basically indexing the (new) sliced array, re-indexing the (original) array after the splice, indexing a (new) array after the concat, and re-indexing the original again when I set it equal to the concat. This seems like it is too inefficient. Is there a more established approach to this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-16T12:54:45+00:00Added an answer on June 16, 2026 at 12:54 pm

    You don’t need to slice() and then splice(). Splice() returns the removed elements, so you just need to do that:

    var n = array.splice(0, 20);
    a2 = array.concat(n);
    

    To be completely clear, JavaScript’s splice() method returns the removed elements, not the remaining elements.

    Also, using globals is generally a bad idea, but you are also kinda mixing them in a weird way. If you are going to keep the variable global, I would either pass the original in as a parameter and return the result from the function:

    var a2=[the array with 80 elements];
    a2 = run(a2);
    
    function run(array){
        var n = array.splice(0, 20);
        return array.concat(n);
    }
    

    OR

    don’t pass it in at all and just reference the global from the get-go:

    var a2=[the array with 80 elements];
    run();
    
    function run(){
        var n = a2.splice(0, 20);
        a2 = a2.concat(n);
    }
    
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