What is the difference between copy and mutableCopy when used on either an NSArray or an NSMutableArray?
This is my understanding; is it correct?
// ** NSArray **
NSArray *myArray_imu = [NSArray arrayWithObjects:@"abc", @"def", nil];
// No copy, increments retain count, result is immutable
NSArray *myArray_imuCopy = [myArray_imu copy];
// Copys object, result is mutable
NSArray *myArray_imuMuta = [myArray_imu mutableCopy];
// Both must be released later
// ** NSMutableArray **
NSMutableArray *myArray_mut = [NSMutableArray arrayWithObjects:@"A", @"B", nil];
// Copys object, result is immutable
NSMutableArray *myArray_mutCopy = [myArray_mut copy];
// Copys object, result is mutable
NSMutableArray *myArray_mutMuta = [myArray_mut mutableCopy];
// Both must be released later
copy and mutableCopy are defined in different protocols (NSCopying and NSMutableCopying, respectively), and NSArray conforms to both. mutableCopy is defined for NSArray (not just NSMutableArray) and allows you to make a mutable copy of an originally immutable array:
// create an immutable array
NSArray *arr = [NSArray arrayWithObjects: @"one", @"two", @"three", nil ];
// create a mutable copy, and mutate it
NSMutableArray *mut = [arr mutableCopy];
[mut removeObject: @"one"];
Summary:
mutableCopy to be mutable, regardless of the original type. In the case of arrays, the result should be an NSMutableArray.copy to be mutable! copying an NSMutableArray may return an NSMutableArray, since that's the original class, but copying any arbitrary NSArray instance would not.Edit: re-read your original code in light of Mark Bessey's answer. When you create a copy of your array, of course you can still modify the original regardless of what you do with the copy. copy vs mutableCopy affects whether the new array is mutable.
Edit 2: Fixed my (false) assumption that NSMutableArray -copy would return an NSMutableArray.