eloquent javascript exercise 4.2 reversing an Array

Question

In the following code I do not understand why reverseArrayOne does not return the reversed array as compared to reverseArrayTwo . In essence I believe I'm assigning the reversedArray to Array in both cases. link to question http://eloquentjavascript.net/04_data.html#c_F3JsLaIs+m

function reverseArray(array) {
reversedArray = [];
  for (i=array.length-1; i>=0; i--) {
    reversedArray.push(array[i]);
  }
  return reversedArray;
}

function reverseArrayOne(array) {
  array = reverseArray(array);
  return array;
}

function reverseArrayTwo(array) {
  reversedArray = reverseArray (array);
  for (i=0; i<reversedArray.length; i++) {
    array[i] = reversedArray[i];
  }
  return array;
}
var arrayValue = [1, 2, 3, 4, 5];
reverseArrayOne(arrayValue);
console.log(arrayValue);
// → [1,2,3,4,5]
reverseArrayTwo(arrayValue);
console.log(arrayValue);
// → [5, 4, 3, 2, 1]

Answer 1

When you pass an array into a function, the function may change the value of the array itself.

reverseArrayOne(arrayValue) returns a reversed array, but does not change the value of arrayValue.

reverseArrayTwo(arrayValue) will actually change the value of arrayValue because of this line in the function:

array[i] = reversedArray[i];

In your log, you are displaying the value of arrayValue instead of the returned result of the function

console.log(arrayValue)

The first function does not change the array value so it'll return [1,2,3,4,5]. The second function actually changes the array value so it'll return [5,4,3,2,1]. If all you need is to print a reverse of an array, then you should do

result = reverseArrayOne(arrayValue);
console.log(result)

Answer 2

Bear in mind the difference between how and object is treated different from values,variables and literals values.I've added comments explaining the procedure,bump the answer incase you need more detail.

a major distinction is: while objects are passed by reference,the latter are passed by value

function reverseArray(array) {
reversedArray = [];
  for (i=array.length-1; i>=0; i--) {
    reversedArray.push(array[i]);
  }
//return a new array with elements in a reverse order of original array
  return reversedArray;
}

function reverseArrayOne(array) {
//scope of object reference is in the function only and hence is not reflected outside
  array = reverseArray(array);
//return the same object pointer which now points to a new array
  return array;
}

function reverseArrayTwo(array) {
      reversedArray = reverseArray (array);
      for (i=0; i<reversedArray.length; i++) {
//you are changing the properties of the same object(this is different from merely changing the reference of the object)
        array[i] = reversedArray[i];
      }
      return array;
    }
    var arrayValue = [1, 2, 3, 4, 5];
//passing arrayValue as reference
    reverseArrayOne(arrayValue);
//original arrayValue still points to the original array since the reverse value is not in scope.
    console.log(arrayValue);
    // → [1,2,3,4,5]
    reverseArrayTwo(arrayValue);
//the original array is the same,but the internal property of the object has been changed and hence is reflected
    console.log(arrayValue);
    // → [5, 4, 3, 2, 1]

Answer 3

I think that can be a good way else...

var list = ["orange", "apple", "pear"]
var newList = [];

function newReverse(list){
  for(var i=0; i<list.length; i++){
    newList.push(list[i]);
  }
  newList.reverse();
  return newList;
}