Contrasting Results When Reversing an Array

Question

Contrasting Results When Reversing an Array

Currently, I am delving into my Javascript book and attempting to master the art of reversing an array by creating a custom reverse function. My challenge is to flip the array without resorting to creating a new variable for storage. After what seemed like finding the solution, running output tests in two different ways (refer below) resulted in contrasting results:

function reverseArrayInPlace(array) {
  for (var i = 0; i < array.length; i += 1) {
    array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
  }
  console.log(array);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

The outputs are as follows:

reverseArrayInPlace(array);
console.log(array);
> [ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 ]
> [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]

Interestingly, I realize that utilizing console.log() inside the function returns the intended result.

However, when I utilize console.log() outside the function, it inexplicably causes the original array configuration with the last element omitted. An explanation regarding this peculiar behavior would be greatly appreciated.

javascript arrays reverse

Answer 1

Answer №1

The array within the function exists within its own scope separate from the global / window level. The global array remains untouched; instead, the function modifies a local copy of it.

If you fail to pass array as an argument to the function, it would affect the globally defined array variable:

function reverseArrayInPlace() { // <-- no parameter
  for (var i = 0; i < array.length; i += 1) {
    array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
  }
  console.log(array);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
reverseArrayInPlace();
console.log(array);

(...though it is generally discouraged to rely on globals in this manner because of the risk of inadvertently masking them with local variables, as demonstrated here. A more recommended approach is for functions to accept data as parameters and return values, allowing you to assign the returned value as needed when calling the function.)

Answer 2

The array within the function exists within its own scope separate from the global / window level. The global array remains untouched; instead, the function modifies a local copy of it.

If you fail to pass array as an argument to the function, it would affect the globally defined array variable:

function reverseArrayInPlace() { // <-- no parameter
  for (var i = 0; i < array.length; i += 1) {
    array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
  }
  console.log(array);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
reverseArrayInPlace();
console.log(array);

(...though it is generally discouraged to rely on globals in this manner because of the risk of inadvertently masking them with local variables, as demonstrated here. A more recommended approach is for functions to accept data as parameters and return values, allowing you to assign the returned value as needed when calling the function.)

Answer 3

Answer №2

When working inside the reverseArrayInPlace function, you are actually reassigning the variable for the array, rather than mutating it. This means that the original array you pass in remains unchanged. The console.log statement inside the function will show the modified array, while the one outside will display the original array.

If you're looking for a different approach, consider using the following code instead:

function reverseArrayInPlace(array) {
  for (var i = 0; i < array.length; i += 1){
    array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
  }
  return array;
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

var newArray = reverseArrayInPlace(array)

console.log(newArray)

Answer 4

When working inside the reverseArrayInPlace function, you are actually reassigning the variable for the array, rather than mutating it. This means that the original array you pass in remains unchanged. The console.log statement inside the function will show the modified array, while the one outside will display the original array.

If you're looking for a different approach, consider using the following code instead:

function reverseArrayInPlace(array) {
  for (var i = 0; i < array.length; i += 1){
    array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
  }
  return array;
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

var newArray = reverseArrayInPlace(array)

console.log(newArray)

Answer 5

Answer №3

Approaching the array reversal process in a unique way, this method involves utilizing a function stack to store array items. The approach includes popping the value, checking the array length, recursively calling the function with the same object reference, and then unshifting the temporarily stored value back into the array.

It's worth noting that while this method may be useful for educational purposes or for small arrays, its reliance on the stack means it may not be suitable for larger datasets due to limitations on stack size.

function reverseArrayInPlace(array) {
    var temp = array.pop();
    if (array.length) {
        reverseArrayInPlace(array);
    }
    array.unshift(temp);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

reverseArrayInPlace(array);

console.log(array);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 6

Approaching the array reversal process in a unique way, this method involves utilizing a function stack to store array items. The approach includes popping the value, checking the array length, recursively calling the function with the same object reference, and then unshifting the temporarily stored value back into the array.

It's worth noting that while this method may be useful for educational purposes or for small arrays, its reliance on the stack means it may not be suitable for larger datasets due to limitations on stack size.

function reverseArrayInPlace(array) {
    var temp = array.pop();
    if (array.length) {
        reverseArrayInPlace(array);
    }
    array.unshift(temp);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

reverseArrayInPlace(array);

console.log(array);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 7

Answer №4

Within the initial loop of your for statement, you are utilizing array.pop() to alter the array provided as an argument. However, afterwards, you are creating a new array and storing it in the same variable, causing the original array's reference to be lost. As a result, the subsequent loops generate the modified array within your function.

To address this issue, I have included a line of code that copies the original array instead of modifying the original argument.

function reverseArrayInPlace(array) {
array = array.slice(); // making a copy of the original array
for (var i = 0; i < array.length; i += 1) {
array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
}
console.log(array);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

reverseArrayInPlace(array);
console.log(array);

Answer 8

Within the initial loop of your for statement, you are utilizing array.pop() to alter the array provided as an argument. However, afterwards, you are creating a new array and storing it in the same variable, causing the original array's reference to be lost. As a result, the subsequent loops generate the modified array within your function.

To address this issue, I have included a line of code that copies the original array instead of modifying the original argument.

function reverseArrayInPlace(array) {
array = array.slice(); // making a copy of the original array
for (var i = 0; i < array.length; i += 1) {
array = array.slice(0, i).concat(array.pop()).concat(array.slice(i));
}
console.log(array);
}

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

reverseArrayInPlace(array);
console.log(array);

Contrasting Results When Reversing an Array

Answer №1

Answer №2

Answer №3

Answer №4

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