Why does fast enumeration not skip the NSNumbers when I specify NSStrings?

Question

I thought that I knew how to use fast enumeration, but there is something I don't understand about it. If I create three NSString objects and three NSNumber objects and put them in an NSMutableArray:

NSString *str1 = @"str1";
NSString *str2 = @"str2";
NSString *str3 = @"str3";

NSNumber *nb1 = [NSNumber numberWithInt:1];
NSNumber *nb2 = [NSNumber numberWithInt:2];
NSNumber *nb3 = [NSNumber numberWithInt:3];

NSArray *array = [[NSArray alloc] initWithObjects:str1, str2, str3, nb1, nb2, nb3, nil];

then I make do fast enumeration on all NSString objects, like this:

for (NSString *str in array) {
    NSLog(@"str : %@", str);
}

In the console, I get this result :

2011-08-02 13:53:12.873 FastEnumeration[14172:b603] str : str1
2011-08-02 13:53:12.874 FastEnumeration[14172:b603] str : str2
2011-08-02 13:53:12.875 FastEnumeration[14172:b603] str : str3
2011-08-02 13:53:12.875 FastEnumeration[14172:b603] str : 1
2011-08-02 13:53:12.876 FastEnumeration[14172:b603] str : 2
2011-08-02 13:53:12.876 FastEnumeration[14172:b603] str : 3

I logged only the NSStrings, but I get a line for every object in the array, even the NSNumbers and I don't understand why. Does fast enumeration always use every object contained in an array?

score 10 · Accepted Answer · answered Aug 02 '11 at 21:13

10

When you write a forin loop like that, it casts every object in the array as an NSString, then prints them out as requested.

If you want only the NSStrings, you would need to write something like this:

for (id obj in array) {
    if ([obj isKindOfClass:[NSString class]]) {
        NSLog(@"str: %@", obj);
    }
}

answered Aug 02 '11 at 21:13

Katfish

698
7
13

More precisely, it casts the *pointer* to be an NSString *pointer*. The object itself is unaffected. – Chuck Aug 02 '11 at 21:24
Casting doesn't exist in Objective-C, only in C. An object thus can't be casted to another class. The type given in the fast enumeration doesn't change anything, every object is passed in. However, `NSLog` does "cast" using the object's `description` method. `NSString` just returns the string, `NSNumber` returns the string equivalent of the number. – Constantino Tsarouhas Feb 28 '12 at 23:35

score 8 · Answer 2 · answered Aug 02 '11 at 21:10

8

The for all loop doesn't know the difference between NSStrings and Integers -- it will simply go through the entire array, cast each as an NSString, and print them out as you asked.

answered Aug 02 '11 at 21:10

Chris Gregg

2,376
16
30

The only reason the code does not crash is that NSLog does not care if you are passing it an NSNumber or NSString when printing. – Kendall Helmstetter Gelner Aug 02 '11 at 21:13
@Kendall Helmstetter Gelner - Good point. Do you know if the values get cast in some way or another? – Chris Gregg Aug 02 '11 at 21:15
@Chris Gregg: If by "the values," you mean the objects themselves, no, there's no such thing as casting an object in Objective-C. Remember, you never actually deal with object values directly in Objective-C. You refer to objects through pointers. Casting from one pointer type to another doesn't change the memory being pointed to — it just tells the compiler to assume there's a different type at the pointed-to address. – Chuck Aug 02 '11 at 21:20
@Chuck - got it. I assume, then, that `NSLog()` has to first figure out what the class of the object is before it prints it to the console? I guess that is true whenever you use the `%@` format; correct me if I'm wrong, but under the covers whatever function is utilizing `%@` will first call `NSObject`'s `- (NSString *)description` method? – Chris Gregg Aug 02 '11 at 21:26
@Chris Gregg: Actually, `NSLog()` doesn't need to know anything about the argument beyond the fact that it's an object — as you correctly note, `NSLog()` just sends `description` to the object and prints whatever string that returns. That process doesn't require `NSLog()` or the argument to know anything about each other; they just need to know that `description` returns a string. – Chuck Aug 02 '11 at 22:29

score 7 · Answer 3 · answered Aug 02 '11 at 21:12

7

I'm pretty sure that fast enumeration returns all objects in the array- all that you're doing in for (NSString *str in array) is typecasting str to an NSString. In the body of the loop you need to check the class of the returned object to make sure that it is an NSString.

for(NSString *str in array)
{
    if([str isKindOfClass:[NSString class]])
        NSLog(@"str : %@", str);
}

answered Aug 02 '11 at 21:12

Carter

3,053
1
17
22

You're completely right. Too bad your answer has lower votes than the wrong answers… – Constantino Tsarouhas Feb 28 '12 at 23:37
@Kiwiiz'Apps Can you mark this answer as the right one? It would help a lot of people. ;-) – Constantino Tsarouhas Feb 28 '12 at 23:40

sergio · Answer 4 · 2012-02-29T15:31:26.663

1

for (NSString *str in array) {

is a way to enumerate through all the elements in array.

You expectative that by specifying NSString you get only the objects of that type is not correct. Rather, all the objects pointers are cast to that type (NSString*).

Have a look at Fast Enumeration in The Objective-C Programming Language guide.

edited Feb 29 '12 at 15:31

answered Aug 02 '11 at 21:10

sergio

68,819
11
102
123

Casting doesn't exist in Objective-C, only in C. An object thus can't be casted to another class. The type given in the fast enumeration doesn't change anything, every object is passed in. However, `NSLog` does "cast" using the object's `description` method. `NSString` just returns the string, `NSNumber` returns the string equivalent of the number. – Constantino Tsarouhas Feb 28 '12 at 23:39
@Randy Marsh, thanks for your comment. Indeed, what I meant is that the pointer is cast from `id` to `NSString*`; only, my wording was not proper. – sergio Feb 29 '12 at 15:34

score 1 · Answer 5 · answered Aug 02 '11 at 21:18

Objective-C is dynamically typed, meaning that at runtime (when the loop actually runs), objects are all effectively one type (id) with different classes. The language allows optional compile-time static typing, but all that does is check whether the messages you're sending are valid for the type you've marked. It doesn't actually change the behavior of your program. If you cast an object to be a different type than it actually is, all you're doing is lying to the compiler and defeating its type-checker.

score 1 · Answer 6 · answered Aug 03 '11 at 07:13

Every object that descends from NSObject implements the method - (NSString)description, %@ in Objective-C formate string will take the corresponding argument for the %@ and call its description method, Most subclasses of NSObject will implement there own version of - (NSString)description. The same thing happens when you type

> po anObject

in the debugger.

score 0 · Answer 7 · answered Feb 05 '13 at 08:27

0

in fast enumeration no typecasting,just assigning the pointer into new object

answered Feb 05 '13 at 08:27

appsign

1

score 0 · Answer 8 · answered Aug 02 '11 at 21:10

I don't understand where is the unexpected behavior, using the enhanced for loop in an NSMutableArray will just iterate thru every single object in the array which in your case is 6, the result is correct and expected.

The numbers will just get casted to Strings.

Why does fast enumeration not skip the NSNumbers when I specify NSStrings?

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