NSString *myString = @"sample string";
NSString *newString = [myString copy];
If I set a breakpoint after these two lines, the pointer for myString is the same as the pointer for newString.
WTF? Isn't NSString copy supposed to return a pointer to a new object? Or am I missing something fundamental about how copy is supposed to work?
Since NSString
is not mutable it might just internally increase ref count
and be done with it.
When you release
one of those NSStrings it might just decrement ref count
- standard memory management.
Do you see any issues with that?
You can Allocate new variable like in sample
NSString *anotherString = [[NSString alloc] initWithString:originalString];
Think about this: NSMutableString
is a subclass of NSString
. When your property is declared as NSString
, you don't expect it to change.
Consider, if you used retain
and someone gave you an NSMutableString
and then later on does change it, your class will be broken.
However, you may think that always copy
ing is slow. So NSString
's copy
simply calls retain
. NSMutableString
's copy
makes an actual copy.
It is usually better to give spit out an NSString *
because people won't have to copy it all the time.
It's a better practice to do copy the string value returned by a method, since the returned value maybe a mutable string object, and this value can be modified in other thread after returned by that method.
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