+ 2

Why is "x.reverse()" reversing var y in the code below?

var x = "hello".split(""); var y = x; x.reverse(); console.log(y); console.log(x); /*i am trying to make a palindrome checker. x and y are both reversed and not just x. why? any way around this? PS: running this code on node.js console*/

13th Apr 2021, 1:15 PM
Mons Joseph
Mons Joseph - avatar
6 Respostas
+ 2
Mons Joseph This is a very good question and it can be a complex topic when you get into the detail, but CarrieForle has given a really good explanation above. If you're interested in some more detail as to why this happens, there's a distinction between primitive types (e.g. string) and reference types (e.g. arrays). When a piece of data that's a reference type (like your array), is assigned to a variable or property, it's not the actual data that gets assigned. It's a reference to that data which gets assigned. When you have 2 or more variables or properties all holding that same data, they are holding references to the data, not copies of the data. Those references all point to the same location in memory. So, if you change the data in one place, it changes everywhere. It's to do with how memory works and saving computational resources. Look up primitive types and reference types (also known as object types). Then you'll get into the domain of pass by reference, pass by value, immutability etc.
13th Apr 2021, 4:22 PM
CamelBeatsSnake
CamelBeatsSnake - avatar
+ 4
Because var x = "hello".split(""); Was stored in the y variable.
13th Apr 2021, 1:23 PM
Matias
Matias - avatar
+ 3
Because array is reference type. When you modify the original array, the assigned one is the reference of the original one and therefore is also modified. http://dyn-web.com/javascript/arrays/value-vs-reference.php
13th Apr 2021, 2:14 PM
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äœ çŸ„é“èŠć‰‡ïŒŒæˆ‘äčŸæ˜Ż - avatar
+ 2
https://code.sololearn.com/W23P2K3naMsK/?ref=app i hope this will help you
14th Apr 2021, 8:21 AM
sami
sami - avatar
+ 1
Just to add a bit of code to what I said above, this shows what happens with primitive types like strings: var string = "code"; var otherString = string otherString = otherString.toUpperCase(); console.log(otherString); //changed to "CODE" console.log(string); // still "code" So, even though we've taken a copy of the first string, it only "changes" in one place. This isn't really a changed string. It's a new string because primitive types can't be changed (immutable).
13th Apr 2021, 4:49 PM
CamelBeatsSnake
CamelBeatsSnake - avatar
+ 1
This problem is solved with the slice() array method: y = x.slice(); And this happens because you assign to the variable "y" not the value of the variable "x", but a link to the array in which the value of the array "x" is stored, (something like this)â˜ș
14th Apr 2021, 2:26 AM
Solo
Solo - avatar