Does the assignment operator make a shallow copy of an NSArray?

Question

Say I have two arrays:

NSArray *A = @[@"Red", @"Blue", @"Yellow"];
NSArray *B = A;

Is B technically a shallow copy of A? If I make changes to the data contained in A, B sees those same changes, and vice-versa. When looking at copying in Apple's documentation, simple equality via the "=" operator is not mentioned as a valid way to make a shallow copy. If this doesn't constitute a shallow copy, then what is it?

Answer 1

Assignment (in Objective-C) is not a copy at all. That is, it's not a copy of the object, it's only a copy of the reference to the object. There's a reason every object (reference) has to be declared as a pointer. (Blocks are a special case.)

A shallow copy is done via the copy method, as in: NSArray *B = [A copy];

As it happens you can't modify NSArrays, but the principle holds for NSMutableArrays as well.

NSMutableArray *a = [@[@"Red", @"Blue", @"Yellow"] mutableCopy];
NSMutableArray *b = a;
NSMutableArray *c = [a mutableCopy];

NSLog(@"%@, %@, %@", a[0], b[0], c[0]);  // Prints: Red, Red, Red

b[0] = @"Green";

NSLog(@"%@, %@, %@", a[0], b[0], c[0]);  // Prints: Green, Green, Red

c[0] = @"Purple";

NSLog(@"%@, %@, %@", a[0], b[0], c[0]);  // Prints: Green, Green, Purple