What happens to a Redis data store if the data exceeds available ram?
What does Redis do when it runs out of memory?
Is the 64-Bit version of Ubuntu only compatible with AMD CPUs? Why do some lenses focus past infinity? On the kit 18-55 lens that came with my Nikon D5100 I have noticed that when trying to focus on infinity for shots in dark conditions (star-trails, etc) the lens tends to move past infinity out of focus and need to be brought back a little to get a clear infinity focus point.  Is there a good trick for getting a clear infinity focus when in pitch dark conditions? I am aware    but I ask this regarding SE-2.0 sites, as the example I will take gaming:  Both answers on the question  make perfect sense to me, i.e. on the one hand this might overuse the tagging system but on the other hand it would help a lot for filtering.  Therefore I suggest a system which, in this case, would internally store the platforms to which a tag is relevant and   if a question is tagged with one of these platforms, nothing special will happen and the question is to be considered platform specific (e.g. a bug which only appears on the PS3 version) if the question is not tagged with a platform, the only result will be that the tag filter mechanism works a bit different compared to what would happen if the question was (invisibly) tagged with all platform tags:   If the invisible tagging is definite on interesting or ignore (i.e. there is not at least one ignored and one interesting tag) the usual mechanism works. if the tagging is ambiguous (i.e. contains both interesting and ignored tags), no filtering happens at all    The actually attributed tags must of course override this behaviour. That way questions about a game one likes but which is available for a platform one doesn't care about will still be highlighted unless specifically tagged with that platform, but no additional tags have to be used to filter a game which is only available on ignored platforms.  This can of course easily be extended to other sites, e.g. the  may be interesting both general and for c++ but totally uninteresting if the question is only about a JavaScript implementation.  I am only unsure whether search should include those hidden tags, too. Proof of property of abelian group Can we use an Android phone as a Wi-Fi repeater? Can we use an Android phone as a Wi-Fi repeater?  Imagine situation when you can access a weak Wi-Fi signal from a free hotspot (coffee shop) signal near window, but want to work in the middle of the room. Then we can put the phone on the window sill to spread the signal over the entire home.  Is it feasible? Instant e-mail notifications of answers to questions I'd like to have Instant notifications by email of answers to my questions.  Can this be implemented? Too many requests? I went to stackexchange.com and clicked on SO.  After a few minutes I got a message saying "We have detected too many requests from this IP.  If this is an error contact us."  That was the second SE page I tried to visit.  Second only to SE.com  I moved to another machine and it's working now.  The first machine usually works, I don't know if a got assigned a blacklisted IP or what.  Anyone know what's going on?  EDIT: I would really rather not post my IP on the web, but if a diamond or dev asks I'll email it.  EDIT 2: I looked at badp's dupe suggestion and can verify that I have no scripts frequently calling SO. I'm using the following command to embed images in my LaTeX document (which is then rendered as PDF)  \newcommand{\img}[4]{   \begin{figure}[!hbt]     \centering       \vspace{1ex}       \includegraphics[width=#2]{images/#1}       \caption[#4]{\label{img.#1} #3}     \vspace{1ex}   \end{figure} }   When creating a href to an image, the viewer jumps to the image caption instead of the top of the image. How do I change my \img command so it still puts the caption below the image but links to it jump to the image itself? How to show that $\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i}=0 $? Show that  $$\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =0 $$  I've proved that this sequence converges (it is bounded and decreasing). NOW, I need to find a sequence that is bigger than this one and goes to zero. Maybe something using geometric serie of 1/2  Thanks in advance! FIELD_INTEGRITY_EXCEPTION: field integrity exception: unknown (invalid user or group: 005XXXXXXXXXXXX) I've a object called myobject__c. This object is private in OWD Setting.  Now when admin create a new record of this object then I want to share this record with a user(community user). For this I've written a code in trigger to create a sharing record of this object (myobject__Share).  But I am getting a problem of :-      FIELD_INTEGRITY_EXCEPTION: field integrity exception: unknown (invalid   user or group: 005XXXXXXXXXXXXX).   how can I solve this problem and second doubt is can we create share record for community user or not. How to prove that this set is closed? Suppose $Y$ be an ordered set in the order topology. Let $f, g: X \to Y$ be continuous. How to show that the set $\{x| f(x) \leq g(x)\}$ is closed?  This is a excercise from munkres. Maybe trying to show that the set $f(\{x| f(x) &gt; g(x)\})$ is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint?  Thanks. Why are the longer edges being beveled more? I'll preface this by saying I'm very new to Blender - I'm trying to put a bevel on an edge, looks something like this:    As you can see, the vertical side is much longer than the horizontal side.  When I apply to bevel to the edge there, the bevel stretches proportionally - so there's 'more' bevel on the longer side.  I want to bevel without any 'ratio' like that - just a 45-degree bevel, if that makes any sense.  Do I have to put in a loop cut to make the side lengths equal?  I tried using a sub-surf modifier, but it applies on a bunch of nearby edges, when I just want to round this one corner. How do the review limits work? I got the brand new Custodian badge today for reviewing posts. Inspired by that, I reviewed some more suggested edits. Now I ran into a limit:     You have no more suggested edit votes today; come back in x hours.   How is that limit calculated? I made 61 reviews today which seems to be some calculated limit.  I searched for an answer on meta but found nothing helpful. Why do my shortcut icons get broken on Windows 7? I have a strange problem. My icons seem to disappear randomly on my Windows 7. They were there a while back, now they ain't.  It happens often and quite randomly. Anyone know a fix? Cleaning up factor levels (collapsing multiple levels/labels) I am reading Velleman's "How to Prove it", and in one example he proves  $\neg \exists x P(x)  \implies \forall x \neg P(x)$    in the following way:    Suppose $\neg \exists x P(x)$. Let x be arbitrary. Suppose $P(x)$. Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x)  \implies \forall x \neg P(x)$    What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?    Edit: I deleted my first example since it was clearly not representative for the problem.  Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does. When, if ever, should voting override answer acceptance for answer sorting? I am using Google Chrome. I re-installed it today as it was crashing when I start it. I guess may be that this was due to any of my Extensions or Google Update. I disabled my extensions but it was still crashing. Now I installed an old version of chrome from . However, Chrome automaticallyuppdates itself (although Windows asks me before installing and I don't grant it permissions, its still annoying to get Google update asking me permission after a couple of minutes.   I saw the question at  however, it doesn't seem to hlp me. i disabled the service and I stopped it as well, but the problem still persists. I also tried to look into the registry according to the following link:  but the "Policies" section of my registry doesn't contain any folder called "Google". I use Windows 7 - 64 bit Home Premium Service pack 1.  Could anybody help? Thanks a ton! How do you clone an Array of Objects in Javascript? I am having trouble loading csv files into layers using QgsVectorLayer().  I know this looks like a repeat, but I have carefully read     and    but still cannot get it work.  An example of my data file, test.csv:  Index,y,x 0,37.7646618,-122.4960858 1,37.777494,-122.416311 2,37.717707,-122.3997056 3,37.732284,-122.498118 4,37.749998,-122.392333   My code:  import os import urllib  import pathlib from qgis.core import QgsVectorLayer  local_path = 'C:\dev\FCMS_conflation\data\TMC_nodes.csv' node_layer = 'test'  abspath = os.path.abspath(local_path) params = {'delimiter': ',',           'decimalPoint': '.',           'xField': longitude,           'yField': latitude}  uri = "%s?%s" %(pathlib.Path(abspath).as_uri(), urllib.unquote(urllib.urlencode(params)))  layer = QgsVectorLayer(uri, node_layer, "delimitedtext")   The formatted URI looks like:  [] uri &gt;&gt;&gt; 'file:///C:/dev/FCMS_conflation/data/TMC_nodes.csv?decimalPoint=.&amp;delimiter=,&amp;xField=longitude&amp;yField=latitude'   I have tried many variations of the parameters, including the crs and type. But no matter what I try, I cannot create a valid layer.  [] layer.isValid() &gt;&gt;&gt; False   I have verified that I can add the file using the QGIS gui. I even tried using the uri provided by layer.metadata from within the Python plugin.   I am using QGIS 2.10.1 on Windows 7 64-bit machine.