Objective-C retain count on assigning object to new pointer

Question

Disclaimer: I'm pretty new to Objective-C

I'm in a command-line project ARC is NOT enabled

I have a class called MyClass

@interface MyClass : NSObject
@end

@implementation MyClass
@end

and my main looks like

int main(int argc, char *argv[])
{
    MyClass *first = [MyClass new];
    MyClass *second = first;
    return 0;
}

Questions:

I know that *first has a retain count of 1. But I don't understand why *second has retain count = 1 as well? I haven't done new, retain, alloc, or copy on that object.
Since *first has retain count of 1, do I have to call release on that? As you can see in the code, I have to release on the object, but Performance Analyzer shows no Memory-leak. Why?
I noticed that both *first and *second has the same value of memory address. Then I suppose that *first retain count should increase to 2 on assigning. But it does not, why?
I noticed that when I retain *first and assign it to *second both of them have retain count of 2 (see below)
```
int main(int argc, char *argv[])
{
    MyClass *first = [MyClass new];
    [first retain];
    MyClass *second = first;
    return 0;
}
```
So basically *second becomes a COPY of *first in that case when is it released?

What happens to the retain count of both pointers if I change the code to

int main(int argc, char *argv[])
{
    MyClass *first = [MyClass new];
    MyClass *second = first;
    [first retain];
    return 0;
}

I mean will *second retain count increase as well?

Thank you in advance.

Note that absolute retain counts are useless when reasoning about memory management. Think in terms of deltas. Www.whentouseretaincount.com has (snarky) details. — bbum
@bbum but they are still essential to understand when learning and debugging. — Justin Meiners
Yes, @justinmeiners, it is essential to understand that the absolute retain count is useless and why (including not being useful for debugging). Because it is useless outside of contrived examples that never happen in real code. — bbum

Justin Meiners Justin Meiners · Accepted Answer · 2013-01-30T15:15:17

You are correct by calling new or alloc an object is created with a retain count of 1.
Assigning second to first does not modify the retain count. Second is simply a pointer to the same object as first. Retain count is a variable on the object so therefore all pointers to the same object will always show same retain count.
You must either assign using a class @property or manually call -retain on to increment the retain count. What you are doing now is called a weak pointer. The assignment of second to first does not modify the original objects retain count.
Yes, once again both pointers are pointing to the same object. Modifications to the object will be reflected in both pointers.
Both pointers are pointing to the same object so by calling [first retain] a single object will be modified and all pointers pointing to it will show an increase in retain count.

You need to think of first and second as pointers, not individual objects. There is only one object in this program and that is the one created at -new. By calling release or retain on either the first or second pointer you are only making changes to the single object pointed to by both.

Objective-C retain count on assigning object to new pointer

1 Answers