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JavaScript JavaScript Foundations Functions Return Values

Posted by Ben Clist
Ben Clist
6,542 Points

Not checking for 'undefined'

      function arrayCounter (arr) {
        if (arr.isArray) {
            return arr.length;
        } else {
            return 0;
        }
      }

I'm getting an error saying that I'm not checking for undefined, but surely if I fed this function undefined it would return 0???

3 Answers

Andrew Shook
Andrew Shook
31,709 Points
Andrew Shook
Andrew Shook
31,709 Points

I have answered this one before here.

Ben Clist
Ben Clist
6,542 Points

Thanks Andrew, this ended up working after I realised my mistake:

      function arrayCounter (arr) {
        if (Array.isArray(arr)) {
            return arr.length;
        } else {
            return 0;
        }
      }
Steven Chin
Steven Chin
2,001 Points
Steven Chin
Steven Chin
2,001 Points

Try this: 

function arrayCounter(arr) {
   if ( ( arr.isArray ) && ( typeof(arr[arr.length]) != 'undefined' ) {
     return arr.length;
   }else {
     return 0;
   }
}
Ben Clist
Ben Clist
6,542 Points
Ben Clist
Ben Clist
6,542 Points

Ignore this, as ever the answer was my terrible attempt at using the isArray.

If anyone else happens to make the same mistake it should be called as Array.isArray(arr).