repeat

Repeat a block of code a fixed number of times

由 自闭症网瘾萝莉.ら 提交于 2019-12-10 03:05:51
问题 I'm trying to repeat a block of code, without using a condition, yet still only repeating it a specific number of times. Basically, something like this: repeat(50) { //Do stuff here. } Is there a way to do this? Other than copying and pasting 50 times? I'm doing this because I figured if I know how many times I want to repeat something, it'd be quicker than checking a condition every time. Is that accurate? Or would I still be checking how many times it's been repeated? Basically, is it any

How to detect if a notification has been dismissed?

由 蹲街弑〆低调 提交于 2019-12-10 01:06:42
问题 Is there any way in Android to detect when a user swipes a notification to the left and deletes it? I'm using an alarmmanager to set a repeating alert and I need my repeating alert to stop when the notification is cancelled by the user. Here's my code: Setting the repeating alert: AlarmManager alarmManager = (AlarmManager) getSystemService(ALARM_SERVICE); alarmManager.setRepeating(AlarmManager.RTC_WAKEUP, c.getTimeInMillis(), repeatFrequency, displayIntent); My notification code: @Override

css start repeating background from defined position

由 …衆ロ難τιáo~ 提交于 2019-12-09 17:46:16
问题 #container{ background:url(images/bg-main.png) repeat-y; width: 903px; background-position: 0px 687px; background-position: bottom; height: 1200px; margin-left: auto; margin-right: auto; } #content{ background:url(images/bg-wood.png) repeat-y; width: 903px; height: 678px; margin-left: auto; margin-right: auto; } #content div is inside #container div. I want #container 's background to start repeating at 687px from top. Is it possible? EDIT: Is it possible that first x pixels of div (from top)

Combinations with repetition

由 爱⌒轻易说出口 提交于 2019-12-09 15:55:54
问题 I'm using Mathematica 7 and with a combinatorica package function I can get all combinations of a certain number from a list of elements where the order doesn't matter and there is no repetition.e.g: in: KSubsets[{a, b, c, d}, 3] out: {{a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}} I cannot find a function that will give me all combinations of a certain number from a list of elements where the order doesn't matter and there is repetition. i.e. the above example would include elements like {a,a,b

Bitmap TileX ONLY

由 限于喜欢 提交于 2019-12-09 08:08:27
I want my image to repeat only in X axis. I use this code: @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); LinearLayout l = new LinearLayout(this); l.setLayoutParams(new LayoutParams(-2, -2)); BitmapDrawable b = null; try { b = new BitmapDrawable(getAssets().open("pattern.jpg")); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } b.setTileModeX(TileMode.REPEAT); l.setBackgroundDrawable(b); setContentView(l); } When i execute my app, bitmap repeat in X axis but i show something string in Y axis http://img24

How Does One Make Scala Control Abstraction in Repeat Until?

由 末鹿安然 提交于 2019-12-09 05:44:15
问题 I am Peter Pilgrim. I watched Martin Odersky create a control abstraction in Scala. However I can not yet seem to repeat it inside IntelliJ IDEA 9. Is it the IDE? package demo class Control { def repeatLoop ( body: => Unit ) = new Until( body ) class Until( body: => Unit ) { def until( cond: => Boolean ) { body; val value: Boolean = cond; println("value="+value) if ( value ) repeatLoop(body).until(cond) // if (cond) until(cond) } } def doTest2(): Unit = { var y: Int = 1 println("testing ...

Generate random number each time and don't include last number

由 我们两清 提交于 2019-12-09 03:46:19
问题 I have 4 colors. I want to make it so that the player can't be the same color 2 times in a row. When a player collides with an object, the RandomColor() is called. So this function is called many times during the game and sometimes the player does not change his color. using UnityEngine; public class ColorManager : MonoBehaviour { public SpriteRenderer player; public string playerColor; public Color[] colors = new Color[4]; private void Awake() { RandomColor(); } public void RandomColor() {

Find longest repeating substring in JavaScript using regular expressions

由 流过昼夜 提交于 2019-12-09 03:16:54
问题 I'd like to find the longest repeating string within a string, implemented in JavaScript and using a regular-expression based approach. I have an PHP implementation that, when directly ported to JavaScript, doesn't work. The PHP implementation is taken from an answer to the question "Find longest repeating strings?": preg_match_all('/(?=((.+)(?:.*?\2)+))/s', $input, $matches, PREG_SET_ORDER); This will populate $matches[0][X] (where X is the length of $matches[0] ) with the longest repeating

How to stop background image repeating for empty space?

由 懵懂的女人 提交于 2019-12-08 20:18:23
问题 Just ran into a problem with repeating background image. In the case when the content is very short, shorter than the monitor height, the background image is still repeating for the extra space. Please refer to the screenshot. Orange bar is my footer. The bottom grey area is the extra space. Can I make stop the background image repeating for the bottom area? I mean ideally the background image just repeats as long as my content. Any help will be appreciated. . 回答1: Rather than sticking the

python regex: capture parts of multiple strings that contain spaces

由 浪子不回头ぞ 提交于 2019-12-08 11:42:26
问题 I am trying to capture sub-strings from a string that looks similar to 'some string, another string, ' I want the result match group to be ('some string', 'another string') my current solution >>> from re import match >>> match(2 * '(.*?), ', 'some string, another string, ').groups() ('some string', 'another string') works, but is not practicable - what I am showing here of course is massively reduced in terms of complexity compared to what I'm doing in the real project; I want to use one
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