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
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
Doesn't javascript uses pointer for arrays ? Should var a = aBackup; do a copy ? otherwise the results seems normal to me...
An array is a reference type of object, so changes made to it will change the underlying value it points to, a and aBackup will point to the same value, and a change made to a will change aBackup aswell.
It is because when you do this, you are not making a copy of the array, but infact a reference to the original array.
var a IS aBackup; // if you will
When you need to do is clone the backup array.
