在foreach循环中消除switch语句

Question

I have question regarding about finding subview using '.tags' in one UIView.

for (UIView *viewObj in [self.view subviews])
{
   // want to find viewObject by its tag.
   // but I heard for-switch or for-case paradigm are bad pattern design by WTF.
   // there are total 9 of tags will be use for finding view object
   switch (viewObj.tag)
   {
         case 0:
         .
         ..
         ...
   }
}

How can I make this to be non-switch or non-if statement? Should I just add into one array then fetch from there? For example, you add views with desired tags and fetch from this array.

Thank you.

Answer 1

From your comments, it's a very finite set of items that will exist - 9 items. If that is expected to be static, there is nothing really wrong with a switch statement. Perhaps you can functionally decompose each switch into a separate method. That said, perhaps the command pattern would be one approach to consider, in addition to your array idea. For reference: http://en.wikipedia.org/wiki/Command_pattern

Answer 2

If it is only going to be a fixed number of views and each view has different methods that need to be performed you could access the views directly.

UIView* someView = [self.view viewWithTag:0];
//Operations on view 0


UIView* someOtherView = [self.view viewWithTag:1];
//Operations on view 1

//...

Answer 3

One thing you could do is have the UIView s in your subviews be of the same type by having each conform to the same protocol. Then you could cast each to id<YourProtocol> and call the same method within each UIView .

Answer 4

为了进一步表达Shan的答案，如果可以将它们分解为函数并将它们（函数指针）放入数组中，则可以使其变得更简单，或者可以使用新的“块”语言功能。

Answer 5

There's a new IBOutlet type, IBOutletCollection. You use that for an NSArray, and then in IB you can add views to that outlet - at runtime, they will all be put into the array (unordered).

To get out a specific one you want, in viewDidLoad you could map all of the NSArray entries into a dictionary keyed by tag value, then just use "objectForKey" to get them out.