Preventing multiple buttons from being touched at the same time

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In iOS is there anyway to prevent a UIView containing multiple buttons (siblings) from being simultaneously from being touched? For instance, two non-overlapping buttons that are side by side can be tapped at the same time with two touches.

Set UIView.exclusiveTouch.

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You can also use below method. If you have two buttons or more, to prevent multiple push at a time.

for e.g,

[Button1 setExclusiveTouch:YES];

[Button2 setExclusiveTouch:YES];

Set this method in your viewDidLoad or viewWillAppear

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for(UIView* v in self.view.subviews)
    {
    if([v isKindOfClass:[UIButton class]])
    {
        UIButton* btn = (UIButton*)v;
        [yourButton setExclusiveTouch:YES];
    }
}

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Swift 4 Syntax:

    buttonA.isExclusiveTouch = true
    buttonB.isExclusiveTouch = true

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You need to find all buttons on that view and set the "exclusiveTouch" property to true in order to prevent multi touch at the same time.

func exclusiveTouchForButtons(view: UIView) {
    for cmp in view.subviews {
        if let cmpButton = cmp as? UIButton {
            cmpButton.exclusiveTouch = true
        } else {
            exclusiveTouchForButtons(cmp)
        }
    }
}

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Comments
  • Awesome! Why I didn't think of that for buttons I do not know?!
  • Perfect ! that was like hidden treasure , never knew of this property, thanks a lot.
  • please note to set it on each "UIButton"! NOT the the UIView those buttons are in :) (Set in on all subviews of that UIView would do) like this -> [self.controlView.subviews makeObjectsPerformSelector:@selector(setExclusiveTouch:) withObject:[NSNumber numberWithBool:YES]];
  • @Hlung: That won't necessarily work; -setExclusiveTouch: takes a BOOL, not a NSNumber (it might work if the number happens to be interpreted as YES, but that's not guaranteed). I call it UIView.exclusiveTouch because that's the class that the property is defined in (a UIButton is a UIView).
  • @Hlung: It is definitely not guaranteed by the docs, which say "The method must take a single argument of type id". AIUI on x86, if the address is divisible by 256 then it will be equivalent to NO (since BOOL is a signed char so it only looks at the bottom 8 bits). I've also heard of UIKit interpreting 2 as NO (i.e. the code only looks at the bottom bit; unsurprising).
  • That answer does not really provide an addition to the already existing answers.