Python

It won't (using your example explicitly) but in general what you've discovered is that:

1. Variables hold values
2. Some of those values are references to shared mutable objects.

Lists fall into the second category. There are ways to copy lists if you want distinct behaviour.

list2 = list1[:]

will perform a "shallow copy". If you have a list of lists, however, the nested lists are still shared references. There is copy.deepcopy available to make a complete clone of something (including all its nested members).

Hum... No it won't.

It's something you only really learn from C or Rust, but your operations will set both to the same list. If you go and change the list, you will change both, but if you set one to a different list, you won't change the other list.

In other words, if you do list1.push(4), you will change list2. But list1 = [3, 2, 1] won't.

Everything's an object. A variable is a pointer to an object. var1 is a pointer to the object 1 (which is an integer). var2 is also a pointer to the same object. If you point var1 to a different object (the integer 2), that doesn't change var2. list1 is a pointer to the list object you defined. list2 is a pointer to the same object. If you change the object, both pointers still point to it.

At least that's my understanding of the whole business. If I got anything wrong, in sure summertime will point it out.

in first case it is pass by value. for list you are passing a reference.

https://www.geeksforgeeks.org/python/pass-by-reference-vs-value-in-python/