Question

Just look at the code and you’ll understand what I mean:

​var aBackup = [3, 4]; // backup array
​​var a = aBackup; // array to work with is set to backup array
a[0]--; // working with array..
a = aBackup; // array o work with will be rested
console.log(a);  // returns [2, 4] but should return [3, 4]
console.log(aBackup);​ // returns [2, 4] too​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​  but should return [3, 4] too

Answer 1

You need to make real copies of your Arrays instead of just using a reference:

var aBackup = [3, 4]; // backup array
var a = aBackup.slice(0); // "clones" the current state of aBackup into a
a[0]--; // working with array..
a = aBackup.slice(0); // "clones" the current state of aBackup into a 
console.log(a);  // returns [3, 4] 
console.log(aBackup); // returns [3, 4]

See MDN for documentation on the slice-method