I'm writing this JavaScript code expecting the output as

[["o","o","x","o","o"],["o","x","x","x","o"]]

but instead it gives

<script>
function createArray(x) {
  var array=[];
  var finalArray=[];
  
  for(var i=1;i<=x;i++){
    array[i-1]='O';
  }

  var midIndex=Math.round(x/2)-1;
  
  array[midIndex]='X';


  finalArray.push(array);

  var num1=midIndex-1;
  var num2=midIndex+1;
 
 
  
    array[num1]="X";
    array[num2]="X";
    
    finalArray.push(array);

  console.log(finalArray) ;
  
}

createArray(5);


</script>

please some one show the reason for this.

I'm writing this JavaScript code expecting the output as

[["o","o","x","o","o"],["o","x","x","x","o"]]

but instead it gives

<script>
function createArray(x) {
  var array=[];
  var finalArray=[];
  
  for(var i=1;i<=x;i++){
    array[i-1]='O';
  }

  var midIndex=Math.round(x/2)-1;
  
  array[midIndex]='X';


  finalArray.push(array);

  var num1=midIndex-1;
  var num2=midIndex+1;
 
 
  
    array[num1]="X";
    array[num2]="X";
    
    finalArray.push(array);

  console.log(finalArray) ;
  
}

createArray(5);


</script>

please some one show the reason for this.

• javascript

• 1 Try finalArray.push(array.slice()); the first time. – Unmitigated Commented Jul 28, 2020 at 6:03
• 1 .push() doesn't clone the array you're pushing into finalArray. You push the "same array" (a reference to the same array) twice. – Andreas Commented Jul 28, 2020 at 6:03
• 2 Is JavaScript a pass-by-reference or pass-by-value language? – Andreas Commented Jul 28, 2020 at 6:06

3 Answers 3

In Javascript, objects(in your case array) are passed by reference. You push the array into finalArray and then update the array. But you have updated the data from same reference. So you could use spread operator which would effectively create a new array(with new reference) and you can push this array to finalArray

finalArray.push([...array]);

That's because Javascript objects are passed by reference. You should create a copy of the array and push to the final one

function createArray(x) {
  var arr=[];
  var finalArray=[];
  
  for(var i=1;i<=x;i++){
    arr[i-1]='O';
  }

  var midIndex=Math.round(x/2)-1;
  
  arr[midIndex]='X';


  finalArray.push(arr.slice());

  var num1=midIndex-1;
  var num2=midIndex+1;
 
 
  
    arr[num1]="X";
    arr[num2]="X";
    
    finalArray.push(arr.slice());

  console.log(finalArray) ;
  
}

createArray(5);

It's for JS pointer When you Push array into final u push pointer of array into an index of final and when change array the final changed too.

for resolve it u can use easy Code:

finalArray.push([...array]);

GoodLuck