CODEWITHSUNDEEP
javascript
BenSV
BenSV

Reputation: 169

JavaScript Arrays .push() not working as Expected

I'm writing this JavaScript code expecting the output as

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

but instead it gives

Received output in the console

<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.

Upvotes: 1

Views: 7558

Answers (3)

Ravi Chaudhary
Ravi Chaudhary

Reputation: 670

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]);

Upvotes: 3

HamidReza Heydari
HamidReza Heydari

Reputation: 300

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

Upvotes: 2

Basheer Kharoti
Basheer Kharoti

Reputation: 4302

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);

Upvotes: 2

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