NSMutableArray arrayWithArray compares same as NSArray mutableCopy?

Question

Given the following code example, is the newMutableArrray variable different depending on the two different initializations, or the same?

NSArray *originalArray = @[obj1, obj2, oj3];

NSMutableArray *newMutableArray = nil;

if (thereIsSomeDifference) {  
    newMutableArray = [NSMutableArray arrayWithArray:originalArray];  
}  
else {  
    newMutableArray = [originalArray mutableCopy];  
}

thereIsSomeDifference is the hypothetical condition for which I would want to initialize newMutableArray one or the other way. This is the essence of the question, because if there is no difference, as your answer indicates, the thereIsSomeDifference condition does not exist. — seeker12
– seeker12, Commented Jan 6, 2013 at 20:46

Johnmph · Accepted Answer · 2013-04-21 10:35:22Z

15

NO !!! There is two differences between these initializations :

Retain count: in the first case, you get an autoreleased object, in the second case, you get a retained object, you will need to release it after (This doesn't apply on ARC)
If originalArray is nil, in the first case, you will get a mutable array with 0 item, in the second case, you will get nil (because sending a message to nil returns nil). In your example, it is clear that originalArray is not nil but in real life you can reach this case (I just had the case)

answered Apr 21, 2013 at 10:35

Johnmph

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Comments

Ramy Al Zuhouri · Accepted Answer · 2013-10-01 23:18:14Z

7

An array is equal to another array (isEqualToArray: selector) if they have the same objects (in the same order). This is verified using the isEqual: method (disregarding of the array being mutable or not).

They're just the same, one or another initialisation doesn't make any difference. Verify this logging the result of isEqualToArray: .

NSArray *originalArray = @[obj1, obj2, oj3];
NSMutableArray *newMutableArray = nil;
newMutableArray = [NSMutableArray arrayWithArray:originalArray];  
thereIsSomeDifference= ![newMutableArray isEqualToArray: [originArray mutableCopy] ];

Notice that the comparison would be true even if you compared it with a non-mutable copy.

edited Oct 1, 2013 at 23:18

answered Jan 6, 2013 at 20:38

Ramy Al Zuhouri

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3 Comments

seeker12 Over a year ago

Thanks for editing to provide additional explanation. Great answer.

rmaddy Over a year ago

You should clarify that two arrays are equal if the objects inside are the same and the objects are in the same order too.

Jawwad Over a year ago

Note: If originalArray is nil, [NSMutableArray arrayWithArray:originalArray] will return an empty NSMutableArray whereas [originalArray mutableCopy] will return nil.

IluTov · Accepted Answer · 2013-01-06 20:36:48Z

1

No, the result of them is exactly the same.
Only the initialisation is different

answered Jan 6, 2013 at 20:36

IluTov

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2 Comments

seeker12 Over a year ago

Thanks for your answer. I accepted @Ramy Al Zuhouri's answer only because it appeared before yours.

IluTov Over a year ago

@seeker12 Mine was first actually, but it doesn't matter, his answer was better.

danh · Accepted Answer · 2013-01-06 20:42:53Z

1

In order to answer, we have to define "sameness". The two inits side by side will result in different collections, but they will be the same insofar as they point to the same elements.

In other words:

initA = [NSMutableArray arrayWithArray:originalArray];  
initB = [originalArray mutableCopy];

if (initA == initB) {
   // unreachable, because the pointers differ
}

// however
if ([initA isEqualToArray:initB]) {
   // will be true
   // because
   for (int i=0; i<initA.count; i++) {
       if ([initA objectAtIndex:i] == [initB objectAtIndex:i]) {
           NSLog(@"this will log every element %@ in the arrays", [initA objectAtIndex:i]);
       }
   }
}

answered Jan 6, 2013 at 20:42

danh

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3 Comments

seeker12 Over a year ago

I suppose "sameness" would be defined as "... they point to the same elements". In what way would the two collections differ?

danh Over a year ago

As I indicated, they would fail a deep equality test a == b.

seeker12 Over a year ago

Ah, ok. I was confused, as my example initializes a single variable one of two ways, based on some condition, and not "...two inits side by side...". Good point. Thanks for the additional explanation.

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