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2 Different Outputs When Reversing Array

I am learning Javascript on a book and have to practice reversing an array by creating my own reverse function. The array must be reversed without creating a new variable to hold the reversed array. I thought I found a solution, but when I try to output my answer in 2 different ways (see below), I get different outputs:

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];


Here are the outputs:

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


When console.log() is used within the function, I get my desired answer.

console.log()

When console.log() is used outside the function, I get the original array with the last element missing. I would like an explanation to this phenomenon.

console.log()

-1

4 Answers
4

The array in the function is on a different scope than that at the global / window level -- the global array is never touched; the function changes a local copy of it instead.

array

array

If you didn't pass array as a parameter to the function, then it would act on the now unmasked global array variable:

array

array

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);

(...although it is generally bad practice to use globals like that, partly because it's easy to accidentally mask them with local variables just as you did here. A better pattern would be for functions to receive their data as params and return a value, so you can decide, when you call the function, what to assign that returned value to.)

Inside reverseArrayInPlace you are reassigning the array variable, not changing (mutating) it. The array you pass in, therefore, is not changed. The inside console.log sees the new array while the one outside sees the original.

reverseArrayInPlace

array

array

Perhaps you want something like this 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)


Just a different approach by using the function stack for storing an item of the array by popping the value, check the length of the array and call the function again with the same object reference and then unshift the temporary stored value.

While working for a small amount of values and because of the limited stack, it is not advisable to use this beside of educational purpose.

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);

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array.length && reverseArrayInPlace(array);

&&

if

In the first loop of your for you are calling array.pop() which modifies the array passed as argument, but then you are creating a new array storing it in the same variable, so the reference to the original array is lost, and then in the subsequent loops the modified array is the one being generated inside your function.

for

array.pop()

Look at this code, I added a line to copy the original array, thus not modificating the original passed as argument.

function reverseArrayInPlace(array)
array = array.slice(); //copying the passed array to not change the original
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);

var a = b

b

a

var a = [1, 2];

a

[1, 2]

var b = a;

a

b

a

b

b

a

array.reverse()

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