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Home/ Questions/Q 757179
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T15:19:57+00:00 2026-05-14T15:19:57+00:00

I always have an class which needs to set up a timer for as

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I always have an class which needs to set up a timer for as long as the object is alive. Typically an UIView which does some animation.

Now the problem: If I strongly reference the NSTimer I create and invalidate and release the timer in -dealloc, the timer is never invalidated or released because -dealloc is never called, since the run loop maintains a strong reference to the target. So what can I do? If I cant hold a strong ref to the timer object, this is also bad because maybe I need a ref to it to be able to stop it. And a weak ref on a object is not good, because maybe i’m gonna access it when it’s gone. So better have a retain on what I want to keep around.

How are you guys solving this? must the superview create the timer? is that better? or should i really just make a weak ref on it and keep in mind that the run loop holds a strong ref on my timer for me, as long as it’s not invalidated?

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  1. Editorial Team
    Editorial Team
    2026-05-14T15:19:57+00:00Added an answer on May 14, 2026 at 3:19 pm

    Treat the lifetime of the timer target as being on screen, not between init and dealloc.

    For UIView do something like this:

    -(void) willMoveToSuperview:(UIView *)newSuperview { [self runMyTimer:newSuperview != nil]; }
    

    For UIViewController do something like this:

    -(void) viewWillAppear:(BOOL)inAnimated { [self runMyTimer:YES]; }
    -(void) viewDidDisppear:(BOOL)inAnimated { [self runMyTimer:NO]; }
    

    Using something like this:

    -(void) runMyTimer:(BOOL)inRun {
      if ( inRun ) {
        if ( !myTimer ) myTimer = [[NSTimer scheduled...] retain];
      } else {
        if ( myTimer ) {
          [myTimer invalidate];
          [myTimer release];
          myTimer = nil;
        }
      }
    }
    

    And for completeness you can call [self runMyTimer:NO] in dealloc.

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