Implement a bottle spin

Implement a bottle spin

By : user3042544
Date : November 28 2020, 12:01 PM
will be helpful for those in need Any time you are tempted to use an undetermined number of variables like name1, name2, it's a big clue that you should not be using individual variables, but should be storing the values in a list like names then use names[0] etc. to access individual items.
In your case you can get the number of players and append to a list in each loop iteration. In the end you'll have a list of player names:
code :
num_players = int(input('How many people are playing?'))
players = []

for i in range(num_players):
    p = input(f'Who is player {i+1}?')


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Does any JVM implement blocking with spin-waiting?

Does any JVM implement blocking with spin-waiting?

By : user3178505
Date : March 29 2020, 07:55 AM
like below fixes the issue Here is evidence that JRockit can use spinlocks - http://forums.oracle.com/forums/thread.jspa?threadID=816625&tstart=494
And if you search for "spin" in the JVM options listed here you will see evidence for the use of / support for spinlocks in Hotspot JVMs.
How to develop a "spin the bottle" application

How to develop a "spin the bottle" application

By : Lou Lembcke
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Hey Check this demo.
You have to calculate velocity. Which depends on touchesMoved:withEvent: and touchesEnded:withEvent:.
Spin bottle with UIGestureRecognizer

Spin bottle with UIGestureRecognizer

By : Joe
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further The simple answer to this is... use a UIScrollView.
From my question here... Loop UIScrollView but continue decelerating
code :
override func viewDidLoad() {
    // Do any additional setup after loading the view, typically from a nib.

    //make the content size really big so that the targetOffset of the deceleration will never be met.

    scrollView.contentSize = CGSize(width: CGRectGetWidth(scrollView.frame) * 100.0, height: CGRectGetHeight(scrollView.frame))
    //set the contentOffset of the scroll view to a point in the center of the contentSize.
    scrollView.setContentOffset(CGPoint(CGRectGetWidth(scrollView.frame) * 50, 0), animated: false)

func rotateImageView() {
    //Calculate the percentage of one "frame" that is the current offset.

    // each percentage of a "frame" equates to a percentage of 2 PI Rads to rotate
    let minOffset = CGRectGetWidth(scrollView.frame) * 50.0
    let maxOffset = CGRectGetWidth(scrollView.frame) * 51.0

    let offsetDiff = maxOffset - minOffset

    let currentOffset = scrollView.contentOffset.x - minOffset

    let percentage = currentOffset / offsetDiff

    arrowView.transform = CGAffineTransformMakeRotation(M_PI * 2 * percentage)

func scrollViewDidScroll(scrollView: UIScrollView) {
    //the scrollView moved so update the rotation of the image

func scrollViewDidEndDecelerating(scrollView: UIScrollView) {
    //the scrollview stopped moving.
    //set the content offset back to be in the middle
    //but make sure to keep the percentage (used above) the same
    //this ensures the arrow is pointing in the same direction as it ended decelerating

    let diffOffset = scrollView.contentOffset.x

    while diffOffset >= CGRectGetWidth(scrollView.frame) {
        diffOffset = diffOffset - CGRectGetWidth(scrollView.frame)

    scrollView.setContentOffset(CGPoint(x: CGRectGetWidth(scrollView.frame) * 50 + diffOffset, y: 0), animated:false)
Rotate UIImage like SPin the Bottle

Rotate UIImage like SPin the Bottle

By : Llakovo
Date : March 29 2020, 07:55 AM
I hope this helps . While it's better if you understand everything yourself, if you want a more concrete example the code here works
Narut's post which is like the 5th response has nice clean code for what you want.
Spin waits, spin loop and busy spin

Spin waits, spin loop and busy spin

By : Martin Hernandez
Date : March 29 2020, 07:55 AM
will be helpful for those in need Busy Spin
A technique which is used in a way that it loop is running until other thread have to complete his work.
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