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Deep copying an NSArray

Is there any built-in function that allows me to deep copy an NSMutableArray?

I looked around, some people say [aMutableArray copyWithZone:nil] works as deep copy. But I tried and it seems to be a shallow copy.

Right now I am manually doing the copy with a for loop:

//deep copy a 9*9 mutable array to a passed-in reference array

-deepMuCopy : (NSMutableArray*) array 
    toNewArray : (NSMutableArray*) arrayNew {

    [arrayNew removeAllObjects];//ensure it's clean

    for (int y = 0; y<9; y++) {
        [arrayNew addObject:[NSMutableArray new]];
        for (int x = 0; x<9; x++) {
            [[arrayNew objectAtIndex:y] addObject:[NSMutableArray new]];

            NSMutableArray *aDomain = [[array objectAtIndex:y] objectAtIndex:x];
            for (int i = 0; i<[aDomain count]; i++) {

                //copy object by object
                NSNumber* n = [NSNumber numberWithInt:[[aDomain objectAtIndex:i] intValue]];
                [[[arrayNew objectAtIndex:y] objectAtIndex:x] addObject:n];
            }
        }
    }
}

but I'd like a cleaner, more succinct solution.

like image 504
ivanTheTerrible Avatar asked Mar 15 '09 03:03

ivanTheTerrible


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You can reverse a NSArray by writing your own loop iterating from the end towards the beginning and using a second array to add the items in reverse order. Or you can simply use - (NSEnumerator *)reverseObjectEnumerator from the NSArray class.

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6 Answers

As the Apple documentation about deep copies explicitly states:

If you only need a one-level-deep copy:

NSMutableArray *newArray = [[NSMutableArray alloc] 
                             initWithArray:oldArray copyItems:YES];

The above code creates a new array whose members are shallow copies of the members of the old array.

Note that if you need to deeply copy an entire nested data structure — what the linked Apple docs call a true deep copy — then this approach will not suffice. Please see the other answers here for that.

like image 75
François P. Avatar answered Oct 03 '22 06:10

François P.


The only way I know to easily do this is to archive and then immediately unarchive your array. It feels like a bit of a hack, but is actually explicitly suggested in the Apple Documentation on copying collections, which states:

If you need a true deep copy, such as when you have an array of arrays, you can archive and then unarchive the collection, provided the contents all conform to the NSCoding protocol. An example of this technique is shown in Listing 3.

Listing 3 A true deep copy

NSArray* trueDeepCopyArray = [NSKeyedUnarchiver unarchiveObjectWithData:
          [NSKeyedArchiver archivedDataWithRootObject:oldArray]];

The catch is that your object must support the NSCoding interface, since this will be used to store/load the data.

Swift 2 Version:

let trueDeepCopyArray = NSKeyedUnarchiver.unarchiveObjectWithData(
    NSKeyedArchiver.archivedDataWithRootObject(oldArray))
like image 36
Andrew Grant Avatar answered Oct 03 '22 06:10

Andrew Grant


Copy by default gives a shallow copy

That is because calling copy is the same as copyWithZone:NULL also known as copying with the default zone. The copy call does not result in a deep copy. In most cases it would give you a shallow copy, but in any case it depends on the class. For a thorough discussion I recommend the Collections Programming Topics on the Apple Developer site.

initWithArray:CopyItems: gives a one-level deep copy

NSArray *deepCopyArray = [[NSArray alloc] initWithArray:someArray copyItems:YES];

NSCoding is the Apple recommended way to provide a deep copy

For a true deep copy (Array of Arrays) you will need NSCoding and archive/unarchive the object:

NSArray *trueDeepCopyArray = [NSKeyedUnarchiver unarchiveObjectWithData:[NSKeyedArchiver archivedDataWithRootObject:oldArray]];
like image 43
Cameron Lowell Palmer Avatar answered Oct 01 '22 06:10

Cameron Lowell Palmer


For Dictonary

NSMutableDictionary *newCopyDict = (NSMutableDictionary *)CFPropertyListCreateDeepCopy(kCFAllocatorDefault, (CFDictionaryRef)objDict, kCFPropertyListMutableContainers);

For Array

NSMutableArray *myMutableArray = (NSMutableArray *)CFPropertyListCreateDeepCopy(NULL, arrData, kCFPropertyListMutableContainersAndLeaves);

like image 41
Darshit Shah Avatar answered Oct 04 '22 06:10

Darshit Shah


No, there isn't something built into the frameworks for this. Cocoa collections support shallow copies (with the copy or the arrayWithArray: methods) but don't even talk about a deep copy concept.

This is because "deep copy" starts to become difficult to define as the contents of your collections start including your own custom objects. Does "deep copy" mean every object in the object graph is a unique reference relative to every object in the original object graph?

If there was some hypothetical NSDeepCopying protocol, you could set this up and make decisions in all of your objects, but unfortunately there isn't. If you controlled most of the objects in your graph, you could create this protocol yourself and implement it, but you'd need to add a category to the Foundation classes as necessary.

@AndrewGrant's answer suggesting the use of keyed archiving/unarchiving is a nonperformant but correct and clean way of achieving this for arbitrary objects. This book even goes so far so suggest adding a category to all objects that does exactly that to support deep copying.

like image 43
Ben Zotto Avatar answered Oct 04 '22 06:10

Ben Zotto


I have a workaround if trying to achieve deep copy for JSON compatible data.

Simply take NSData of NSArray using NSJSONSerialization and then recreate JSON Object, this will create a complete new and fresh copy of NSArray/NSDictionary with new memory references of them.

But make sure the objects of NSArray/NSDictionary and their children must be JSON serializable.

NSData *aDataOfSource = [NSJSONSerialization dataWithJSONObject:oldCopy options:NSJSONWritingPrettyPrinted error:nil];
NSDictionary *aDictNewCopy = [NSJSONSerialization JSONObjectWithData:aDataOfSource options:NSJSONReadingMutableLeaves error:nil];
like image 24
Mrug Avatar answered Sep 30 '22 06:09

Mrug