how to define a returning NSString function in Objective-C / Xcode using a temporary variable?

Question

I would like to define the following function in Objective-C. I have provided pseudo-code to help illustrate what I'm trying to do.

PSEUDOCODE:

function Foo(param) {
  string temp; 

if(param == 1) then
  temp = "x";
else if(param == 2) then
  temp = "y";
else if(param == 3) then 
  temp = "z";
else
  temp = "default";
end if    

  return temp;
}

For some reason if I do this... the variable who I assign it to results in a "BAD Access" error.

I don't know what the difference between:

static NSstring *xx;

or the non-static:

NSString *xx;

declarations are, and how or why I would want to use one over the other.

I also do not fully understand the initializers of NSString, and how they differ. For example:

[[NSString alloc] initWithString:@"etc etc" ];

or the simple assignment:

var = @""

or even:

var = [NSString stringWithString:@"etc etc"];

Can you give me a hand please?

So far, using the NSString value returned from functions like those listed above, always causes an error.

Solution 1

static NSstring *xx;

That declares a statically allocated variable, much like it does in C.

NSstring *xx;

Inside a method that declares a normal local stack variable, just as it does in C.

As you should be aware, the difference between the two is that the first will keep its value between invocations of the function (and can cause trouble if the function is called from multiple threads).

[[NSString alloc] initWithString:@"etc etc"]

That creates a new NSString object, with the contents etc etc. This may or may not be the same as any other NSString object in your program with the same contents, but you don't have to care. Memory management wise, you own it, so you are responsible for ensuring that you eventually call release or autorelease on it to avoid leaking memory.

@"etc etc"
[NSString stringWithString:@"etc etc"]

Those are basically the same. Both give you an NSString object with the contents etc etc. This may or may not be the same as any other NSString object in your program with the same contents, but you don't have to care. Memory management wise, you do not own it, so you must not call release or autorelease on the object unless you first took ownership by calling retain. Also, since you do not own it, you can use it within your method, pass it as a parameter to other methods, and even use it as the return value from your method, but you may not store it in an ivar or static variable without taking ownership by calling retain or making a copy (with copy).

Also, note that "" and @"" are very different. The first gives you a const char * exactly as it does in C, while the second gives you an NSString object. Your program will crash if you use a const char * where the code expects an NSString object.

Solution 2

You can do it this way:

- (NSString *)functionName:(int)param {
    NSString *result = nil;

    switch (param) {
        case 1:
            result = [NSString stringWithString:@"x"];
            break;
        case 2:
            result = [NSString stringWithString:@"y"];
            break;
        case 3:
            result = [NSString stringWithString:@"z"];
            break;
        default:
            result = [NSString stringWithString:@"defaultv"];
            break;
    }

    return result;
}

Solution 3

Post real code, not pseudo code, as it makes it much easier to answer your question in concrete terms.

Given that you indicate that you are quite new to Objective-C, I would suggest starting with the language guide and then moving on to the memory management guide.