CODEWITHSUNDEEP
iosswift3dtouch
Jad Ghadry
Jad Ghadry

Reputation: 265

3D touch equivalence for iPad

I have attempted to develop a game using swift that leverages the 3D touch hardware of iPhones. However, as I was submitting my app to the App Store, it got rejected because the game wasn't playable on iPads.

My question is, what is the best way of implementing a similar functionality for non 3D touch devices? The way I am doing right now is by implementing the following method

override func touchesMoved(_ touches: Set<UITouch>, with event: UIEvent?) {

    if self.didStartGame, let touch = touches.first {

        let maximumPossibleForce = touch.maximumPossibleForce
        let force = touch.force
        let normalizedForce = min(force/maximumPossibleForce * 2.5, 1)

        // Added game-related code here to modify scene accordingly

    }

}

When running the latter on non 3D touch devices, debugging the value of touch.maximumPossibleForce returns 0.

Upvotes: 0

Views: 147

Answers (1)

joern
joern

Reputation: 27620

You cannot detect force touch on devices that that don't support it.

But maybe you could use the majorRadius property on UITouch. It gives you the radius of the touch.

With the radius you can allow users that don't have a 3d touch device to control your game with the angle of their finger:

enter image description here

This is the code for the above example:

override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
    guard let pseudoForce = touches.first?.majorRadius else { return }
    label.text = "\(pseudoForce)"
}

override func touchesMoved(_ touches: Set<UITouch>, with event: UIEvent?) {
    guard let pseudoForce = touches.first?.majorRadius else { return }
    label.text = "\(pseudoForce)"
}

override func touchesEnded(_ touches: Set<UITouch>, with event: UIEvent?) {
    label.text = "-"
}

Upvotes: 2

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