Creating copies in Julia with = operator

问题:

When I create some array A and assign it to B

A = [1:10]
B = A

I can modify A and the change reflects in B

A[1] = 42
# B[1] is now 42

But if I do that with scalar variables, the change doesn't propagate:

a = 1
b = a
a = 2
# b remains being 1

I can even mix the things up and transform the vector to a scalar, and the change doesn't propagate:

A = [1:10]
B = A
A = 0
# B remains being 1,2,...,10

What exactly does the = operator does? When I want to copy variables and modify the old ones preserving the integrity of the new variables, when should I use b = copy(a) over just b=a?

回答1:

The confusion stems from this: assignment and mutation are not the same thing.

Assignment. Assignment looks like x = ... – what's left of the = is an identifier, i.e. a variable name. Assignment changes which object the variable x refers to (this is called a variable binding). It does not mutate any objects at all.

Mutation. There are two typical ways to mutate something in Julia: x.f = ... – what's left of the = is a field access expression; x[i] = ... – what's left of the = is an indexing expression. Currently, field mutation is fundamental – that syntax can only mean that you are mutating a structure by changing its field. This may change. Array mutation syntax is not fundamental – x[i] = y means setindex!(x, y, i) and you can either add methods to setindex! or locally change which generic function setindex!. Actual array assignment is a builtin – a function implemented in C (and for which we know how to generate corresponding LLVM code).

Mutation changes the values of objects; it doesn't change any variable bindings. After doing either of the above, the variable x still refers to the same object it did before; that object may have different contents, however. In particular, if that object is accessible from some other scope – say the function that called one doing the mutation – then the changed value will be visible there. But no bindings have changed – all bindings in all scopes still refer to the same objects.

You'll note that in this explanation I never once talked about mutability or immutability. That's because it has nothing to do with any of this – mutable and immutable objects have exactly the same semantics when it comes to assignment, argument passing, etc. The only difference is that if you try to do x.f = ... when x is immutable, you will get an error.

回答2:

This behavior is similar to Java. A and B are variables that can hold either a "plain" data type, such as an integer, float etc, or a references (aka pointers) to a more complex data structure. In contrast to Java, Julia handles many non-abstract types as "plain" data.

You can test with isbits(A) whether your variable A holds a bit value, or contains a reference to another data object. In the first case B=A will copy every bit from A to a new memory allocation for B, otherwise, only the reference to the object will be copied.

Also play around with pointer_from_objref(A).