Assigning return value of a class method to a variable error

Question

For the last few hours, I have been trying to do something that makes perfect sense to me but not to Xcode, and I still haven't figured out a way to do it. So here it goes.

NSArray *array = [MedInfo getCategoryFirstIndexArray];

Here, getCategoryFirstIndexArray is just a class method of MedInfo:

@interface MedInfo : NSObject
+(NSArray *)getCategoryFirstIndexArray;
@end

But Xcode is keep telling me there is an error in the code, saying 'Expected expression'?? Funny thing is, if I put [MedInfo getCategoryFirstIndexArray]; only, the compiler says it's fine, but as soon as I'm trying to assign the return value to another variable, there is an error.

So what am I doing wrong here? Am I asking too much?

EDIT: Ok, this is dumb. I have been trying to make a new variable inside a switch statement, something like this:

switch(self.viewType.integerValue)
{
    case 0:
        break;
    default:
        NSArray *array = [MedInfo getCategoryFirstIndexArray];
        break;
}

And apparently I didn't know that objective-C doesn't support that (another reason to hate this horrendous language). Sorry for my ignorance, and thank you whoever you are for trying to help! Good bye!

@KyrDunenkoff Try to be helpful without being rude. The need to use curly braces only if the first line of a `case/default` block declares a variable is a pretty subtle bit of knowledge. Finding that bit of info in "the documentation" is not going to be a simple task for anyone. — rmaddy, Aug 11 '13 at 16:11
@rmaddy `s/the documentation/a reasonable beginner's guide/`, Kyr is absolutely right. (Also, OP could have googled the error message. That's what I always did with my beginner errors. Guaranteed instant solution within 10 minutes without wasting others' precious time...) — , Aug 11 '13 at 16:15
@H2CO3 It's still no excuse for being rude. Either help (without being rude) or downvote for lack of effort. — rmaddy, Aug 11 '13 at 16:16
@rmaddy Not a simple task, I agree. What's the excuse for not searching for an answer first? I spent about a year on SO before asking my first question, 'cause 99% of my issues were already answered here. — Kyr Dunenkoff, Aug 11 '13 at 17:09
@KyrDunenkoff I agree there is no excuse for being lazy and not searching first. You have the choice to downvote for lack of effort and/or supply an answer or comment. Either way, there is no need to be so harsh. There are much more polite and constructive ways to point out the lack of effort or duplicate questions. Enjoy. — rmaddy, Aug 11 '13 at 18:40

score 3 · Answer 1 · edited May 23 '17 at 12:20

There are two different problems here.

Neither C nor Objective-C allow the declaration of a variable directly after a label (see Declaring variables inside a switch statement or this answer: https://stackoverflow.com/a/1115323/1187415). So
```
switch (k) {
    case 10:
        break;
    default:
        int x = 20; // ERROR: Expected expression
        break;
}
```
is a syntax error in either language. It works if you add a semicolon after the label:
```
    default: ;
        int x = 20; 
        break;
```
and this works in both C and Objective-C. So one can declare variables inside a switch statement without using brackets { ... }.

With Automatic-Reference Counting this does not work any longer for Objective-C objects (compare When converting a project to use ARC what does "switch case is in protected scope" mean?):

switch (k) {
    case 10: ;
        NSMutableArray *a = [NSMutableArray array];
        // ...
        break;
    default: // ERROR: Switch case is in protected mode
        break;
}

The reason is that the ARC compiler needs a well-defined scope to control the lifetime of the object. And now the only solution is to introduce an additional scope with { ... }:

switch (k) {
    case 10: {
            NSMutableArray *a = [NSMutableArray array];
            // ...
        }
        break;
    default: // Switch case is in protected mode
        break;
}

So this is more an ARC issue than an Objective-C problem.

Assigning return value of a class method to a variable error

1 Answers1