Here's the scenario: I create a rounded rectangle (shape layer) with a border radius of 8px (all sides). Months later I'm ready to code the design only to forget the border radius. Is there any way to rediscover these attributes (aside from properly naming the layer obviously)?
I have a psd that has rounded rectangle. I need to create a new rounded rectangle with the same radius. How do I determine the radius used on that said rounded rectangle?   I feel like this is such an awful question but thanks in advance for answering.
How to move a window if titlebar is overlapped by Ubuntu menu Somehow the title-bar of chrome was overlapped by Ubuntu menu-bar like attached image.  In this case how to move this chrome window? How to replace a string with a string containing slash with sed? LibreOffice Calc - Disable rounding There are a  on this topic, but no simple and definitive answers.    Can someone please tell me how to permanently disable cell rounding on the .csv file which is created by script?  Basically, '377635637222436751' is being converted to '377635637222437000' automatically and that isn't really acceptable for my needs.  I've tried this: Tools->Options->Calc->Calculate "Precision as shown"   and this: Format > Cells > Numbers > Category > (scroll down for) Text. (but considering my script creates the .csv -- this option isn't very good since it must be done after the file is created)  And neither is working.  The number stays rounded -- even after formatting the column after the file is created.  Does anyone have any suggestions? Handle either a list or single integer as an argument Supporting Multiple Screen Sizes in Storyboards (3.5 and 4 inch screens) How do I tokenize a string in C++? Java has a convenient split method:  String str = "The quick brown fox"; String[] results = str.split(" ");   Is there an easy way to do this in C++? Is the sign of principal components meaningless? Does the sign of the vectors returned by PCA have any significance after I center my variables? I am attempting to solve this counting problem through combinatorial argument. The following is the equation I am given:   $$\sum_{i=1}^n (i-1)(n-i) = \binom{n}{3}$$  I understand that the right-hand side of this equation represents a set of $n$-elements out of which we choose 3. For example I believe we can say suppose we have a group of $n$ people and we want to choose 3 out of $n$ to be in a committee. However I'm not sure how to express the left-hand side in words. If forming a committee is an appropriate way to tackle this problem then I know the left-hand side must utilize the addition and multiplication principles, but I don't know how to put it into words. Also my intuition tells me that in solving this we should first flip $$(n-i)(i-1)$$  Thanks! A fair coin is tossed repeatedly until 5 consecutive heads occurs.   What is the expected number of coin tosses? I'm having trouble understanding how to access each individual result returned from a subquery. The picture shows the code which:   takes fields and objects chosen by a user and assigns them to String variables, the string vars are used to query a dynamic query string, the query string is feed into a database.query and the result is stored in an array I can individually access each of the query results from this array (parent object results),  But I have not been able to individually access the subquery results.   If I feed the subquery results into an array...something like:  SObject[] c = p.get(i).getSObjects(childO);   I still can't get to each individual child record per parent. I can only see all child records per each parent. I need to get to each child record per adult, so that I can roll up results.  Can anyone help?  // These strings will be populated dynamically from a VF page String parentO  = 'Account'; String parentF  = 'Products_Purchased__c'; String childO   = 'Opportunity';     if(childO  == 'Opportunity'){ childO = 'Opportunities'; } String childF   = 'Product_Type__c';  // create dynamic SOQL query string String query = 'SELECT id, (select ' +childF+ ' from ' +childO+') FROM ' +parentO;  // query and store 33 results in variable p SObject[] p = Database.query(query);  for(Integer i = 0; i &lt; p.size(); i++){     system.debug('Parent result from SOQL query: ' +p[i]);     if(p.get(i).getSObjects(childO) &lt;&gt; null){         system.debug('Child result(s) from SOQL query: ' +p.get(i).getSObjects(childO));         system.debug('Child field: ' +c.get(i).get(childF));     } } Prefix or adjective meaning "one and a half" Is there a prefix or adjective that means "one and a half", as "tri-" or "triple" is for "three"?  The exact usage I have is to describe "18" in terms of a dozen. Where I live they've started making 18-egg cartons, and the local grocer was wondering what name to give them in his computerised cash register, hoping for a single word.  Even slang, informal or invented term would do if there isn't such a term already. Decomposition of semisimple Lie algebra via its roots I got  question in review. First sentence is:     I know there is a  question, but no answers.   And indeed the two questions are virtually identical. I started to flag as duplicate - but the original question had no answer at the time, so it wouldn't be correct and apparently . (I didn't go so far as to try to submit the flag).  What is the proper action here? How do we want such a question to be handled? Should both (or more) questions stay open indefinitely until either is answered and somebody gets the good idea to do the duplicate linking?   As the new question is essentially a bump to the front page (perhaps with re-tagging), especially in the case where the OP knows that a previous question exist, perhaps something should be done to enable bumping re-emphasizing without writing the new question.  As I wrote this,  actually got an answer. So the old one should now be closed as a duplicate of the new - but my question remains for the case where all known/found duplicate questions have no answer?  Ironically, just before asking this, the "Questions that may already have your answer" gave me ... (Although the suggestion has no answer). So it has been asked before - the comment there even suggest the ability to re-emphasize. How to get rid of update notifications for a given app in Google Play Store? General question I have an app X installed on my device in version Y. I don't want to update it to any newer version (Y+z) for some reason. Still, the Google Play Store app always lists updates for X, which keeps me from a simple &quot;update all&quot;. My question, put to a single sentence:  How to hide updates for X permanently from the &quot;available updates&quot; list?   Specific examples First, I know there are alternatives as e.g. the , which can handle this quite well (skip this update, skip all future updates). I prefer a &quot;generic solution&quot;, which does not depend on a 3rd party's market app. Second, I know that  includes a &quot;Market doctor&quot;, which can be used to disconnect an app from Google Play.1 However, in my specific case that only worked for a few hours: I disconnected GTalk, as I don't want it replaced by Hangouts. The update notification disappeared for a few hours, then it was back. No idea what restored the &quot;link&quot;. Third, just to make sure: I'm not talking about the &quot;auto update&quot; feature as discussed e.g. in . I was rather hoping for something close to , but without the need to &quot;freeze&quot; the app in question (which I still want to use, but in the version installed, not in a &quot;future version&quot; available). Fourth: No, uninstalling the app in question is no option here, though it would perfectly hide any future updates from being displayed :) Nor is &quot;freezing&quot;, as the app is used regularly. Fifth: There is a related question for my GTalk example:  However, I want the opposite: No update.  It might very well be that &quot;Google Apps&quot; (and other pre-installed apps) are handled in some specific way, and restore their &quot;market links&quot; themselves somehow. For this case, my question includes: How to prevent that from happening?  Update There's a  discussing the very same issue. Until recently, a work-around for &quot;user installed apps&quot; was to uninstall them and then sideload the .apk – but even that stopped working. Well, re-signing the .apk with a different key might work for user-apps. But both would fail for pre-installed apps (can't sideload them to override; different signature would refuse that install completely). Moreover, there's an  on this by some &quot;joel.bou..&quot;. Not hard to guess, though blanked out: that would be Joël Bourquard. If it's really him, then that's really an issue (for those who don't know: Joël is the dev behind Titanium Backup). Let me quote from the issue:  For the record: in the Android Market days it was possible to &quot;Attach&quot; or &quot;Detach&quot; apps (on rooted devices) with a system tool such as Titanium Backup. But nowadays (with the Google Play Store app) this has become nearly impossible to do, because the complete app list seems to sync from Google's servers and the user has no control over that.  Which explains why detaching with apps like Titanium Backup is no longer permanent. And I'm afraid I've asked the impossible. But I thought the same in the past with other questions, and surprisingly got solutions here. So I do not yet give up hope on this one!  1: what that does is explained by eldarerathis  (thanks to  for digging that up!) S(x), which stands for “x is a surgeon”  P(x, y), which stands for “x is a patient of y”  R(x), which stands for “x recovers”  H(x), which stands for “x is happy”  1>"A surgeon with no patients is happy" means: for any person xx, if xx is a surgeon and there is no person yy for which yy is a patient of xx, then xx is happy.  2> "A surgeon is happy if all her patients recover" means: for any person xx, if xx is a surgeon and - if for every person yy for which yy is a patient of xx, we have that yy recovers - then xx is happy.  I have these as solution:  1:∀(x) ( S(x)∧ ∀(y)P(y,x) → H(x) ) 2: ∀(x) [ { S(x)∧ (∀(y)P(y,x) ∧ R(y)) } → H(x)]  Are these okay?  And how do prove that 1 follows from 2 by resolution (not by drawing a truth table)? There is a common story that airline ticket prices rise significantly if you buy them less than two weeks before departure.  And that any time before that will yield standard prices.  Is that advice valid for flying during holiday seasons?  As an example: if I want to fly coast to coast USA on December 7th, must I buy my ticket ASAP, or would I get a similar price buying on November 21st? Is a bivariate function that is a polynomial function with respect to each variable necessarily a bivariate polynomial? Composition of two Riemann integrable functions Suppose $f,g$ are two Riemann  Integrable functions .Is it true that $f\circ g$ is also Riemann  Integrable?  Trying this for a long time but not getting the answer Why doesn't the UK ask for travel bookings for a tourist visitor visa? Calculating optimal number of bins in a histogram I'm interested in finding as optimal of a method as I can for determining how many bins I should use in a histogram. My data should range from 30 to 350 objects at most, and in particular I'm trying to apply thresholding (like Otsu's method) where "good" objects, which I should have fewer of and should be more spread out, are separated from "bad" objects, which should be more dense in value. A concrete value would have a score of 1-10 for each object. I'd had 5-10 objects with scores 6-10, and 20-25 objects with scores 1-4. I'd like to find a histogram binning pattern that generally allows something like Otsu's method to threshold off the low scoring objects. However, in the implementation of Otsu's I've seen, the bin size was 256, and often I have many fewer data points that 256, which to me suggests that 256 is not a good bin number. With so few data, what approaches should I take to calculating the number of bins to use?