NSSet of touchesShouldBegin: of UIScrollView is just one NSString

Question

I have a subclass of UIScrollView, it is also the delegate.

When I have the next protocol function called:

- (BOOL)touchesShouldBegin:(NSSet *)touches withEvent:(UIEvent *)event inContentView:(UIView *)view
{
id  element;
NSEnumerator *setEnum =   [touches objectEnumerator];

while ((element = [setEnum nextObject]) != nil)
{
    NSLog(@"element:%@", element);    
}

return [super touchesShouldBegin:touches withEvent:event inContentView:view];
}

The only thing that NSLog is showing is:

element:<UITouch: 0x15a9a0> phase: Ended tap count: 3 window: <UIWindow: 0x392390; frame = (0 0; 320 480); layer = <UIWindowLayer: 0x382cd0>> view: <UIButton: 0x3a5e90; frame = (0 0; 106 138); opaque = NO; layer = <CALayer: 0x3a3e90>> location in window: {228, 126} previous location in window: {227, 126} location in view: {89.6667, 77} previous location in view: {88.3333, 77}

The problem is, it shows the content of the NSSet as one big NSString. If I ask allObjects from the objectEnumerator, I just get one object in an NSArray. Exactly the same object (the NSString) as NSLog is showing.

Could someone please tell me if I am doing something wrong, or if it is not normal that the NSSet is just giving one NSString.

Thank you!

I got confused, it looked like a string indeed. Placing the object from the [objectEnumerator nextObject] in a UITouch variable helped! Thank you all for the quick respones — MiiChiel, Feb 09 '12 at 18:26

zaph · Accepted Answer · 2012-02-09T18:32:07.403

3

The NSLog() output is the result of the description method called on the object. NSLog() output is designed to be human readable for debugging only.

<UITouch: 0x15a9a0> indicates the the object is a UITouch at address 0x15a9a0. The rest of the description just displays some interesting values of the UITouch.

Try:

NSLog(@"tap count:%d", element.count);

etc. and you will se that it is not just a string.

As an aside there is no need for an explicit NSEnumerator, fast enumeration ca be used: you can simplify the code to:

for (UITouch *element in touches)
{
    NSLog(@"element:%@", element);    
}

edited Feb 09 '12 at 18:32

answered Feb 09 '12 at 18:14

zaph

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I understand, but should it not be that when I use NSLog for just one element in the 'while' loop, that I would get only one of all objects been showing in the console? – MiiChiel Feb 09 '12 at 18:16
You are getting a description of one object, the `UITouch` and as a b onus some of it's state. – zaph Feb 09 '12 at 18:25
Great answer man, I kept using `UITouch *touch = [[touches objectEnumerator] nextObject]` and I hate that syntax. – bobobobo Dec 08 '12 at 01:21
When you say "fast enumeration", do you mean it performs better? – bobobobo Dec 08 '12 at 01:21
@bobobobo Fast enumeration means the form of enumeration using a for loop with the keyword "**in**". `for (id obj **in** collection) – zaph Dec 08 '12 at 23:19

score 1 · Answer 2 · answered Feb 09 '12 at 18:18

1

The NSSet touches is a set of UITouch instances, not NSStrings. When you send it to NSLog, it calls the "description" method of the UITouch instance which does return an NSString.

answered Feb 09 '12 at 18:18

picciano

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Thank you that helped, placing the first 'nextObject' in a UITTouch variable got me the object. – MiiChiel Feb 09 '12 at 18:29

NSSet of touchesShouldBegin: of UIScrollView is just one NSString

2 Answers2