memory management in Objective C,NSString - iphone

Question

What I did is
int main(int argc, char *argv[]){
NSString *str = [[NSString alloc] init];
str = #"a a a a a a a a a ";
[str release];
NSLog(#"string is %#",str);
return 0;
}
I think string will be nil because we release it before printing it. However, it is not. Therefore, my consumption is wrong.
Can anybody advice me and correct me. Any comments are welcomed. Thanks

Answer 1

You are leaking memory there, first you allocate a new, empty, NSString object and then you assign a static string literal to the pointer, effectively leaking the memory you previously allocated. Then you release an object that you don't own as pointed out by The Saad and then you should note that release doesn't set the pointer to nil but just releases the object which might or might not end in an de-allocation (which, assuming that it would dealloc the object, would also not set the pointer back to nil).

Answer 2

NSString *str = [[NSString alloc] init];
This declares a variable that is a pointer to an instance of NSString, allocates memory for an instance of NSString, initialises that memory, then assigns a pointer to that memory to the variable you declared.
str = #"a a a a a a a a a ";
This creates an NSString constant and assigns a pointer to it to the variable that you declared. As you no longer have a reference to the original NSString object you created, you have leaked the memory you allocated for it.
[str release];
This declares that this code no longer wishes to be an owner of the memory allocated to str. As you created it with a string literal, this object will always exist in your program, and releasing it does nothing.
I think string will be nil because we release it before printing it.
Releasing an object does not ever set it to nil.

Answer 3

The str variable is a pointer to the string object and essentially contains a memory address. Sending a message like retain to the object does not change the value of the variable.
Just because the string is released will not cause the variable to lose its value. You must expicitly set the variable to nil for that to happen.
Note that with ARC under iOS 5 you can get weak references/pointers which will become nil when their underlying object is released. You do this by prepending a variable declaration with the __weak keyword (or use weak in a property declaration).

Answer 4

String constants are not to be released; they're neither explicitly to be released neither autoreleased; they're simply constants (by the way, they're deallocated at end of scope).
So, only -release them if you previously retained them

Answer 5

Here is the source corrected:
I'm afraid that there is much to learn before fully understanding what is going on. You should also look at understanding malloc (don't use it though), and how pointers work in C.
int main(int argc, char *argv[]){
NSString *str = [[NSString alloc] initWithString:#"a a a a a a a a a "];
[str release];
str = nil;
NSLog(#"string is %#",str);
return 0;
}
int main(int argc, char *argv[]){
NSString *str = #"a a a a a a a a a ";
// DONT Call this - the str [str release]; - no need to release this kind of string
NSLog(#"string is %#",str);
return 0;
}