Iterate a list of types and use it as a generic parameter I am trying to enumerate a list of types in C# and pass these as a generic type. This way I can have one list of object types and use this to sync data, instantiate storage tables, etc. Something like this:  public async Task&lt;bool&gt; SyncAll() {     var types = new List&lt;Type&gt; { typeof(Product), typeof(Customer) };      foreach (var t in types)     {         await Sync&lt;t&gt;();     }     }  public async Task&lt;T&gt; Sync&lt;T&gt;() where T: Syncable {     var items = _repository.GetItems&lt;T&gt;();     ... }   The code above does not work, but is this even possible?
Pass An Instantiated System.Type as a Type Parameter for a Generic Class The title is kind of obscure. What I want to know is if this is possible:  string typeName = &lt;read type name from somwhere&gt;; Type myType = Type.GetType(typeName);  MyGenericClass&lt;myType&gt; myGenericClass = new MyGenericClass&lt;myType&gt;();   Obviously, MyGenericClass is described as:  public class MyGenericClass&lt;T&gt;   Right now, the compiler complains that 'The type or namespace 'myType' could not be found." There has got to be a way to do this.
Einstein's big revelation was that time and space are inseparable components of the same fabric. Physical observation tells us that distant galaxies are moving away from us at an accelerated rate, and because of the difficulty (impossibility?) of defining a coordinate system where things have well defined coordinates while also moving away from each other without changing the metric on the space, we interpret this to mean that space itself is expanding.   Because space and time are so directly intertwined is it possible that time too is expanding? Or perhaps it could be contracting? ASP.NET Call Code Behind Function From Javascript I have a bunch of magnets (one of those game-board thingies) given to me when I was a school-going lad over 20 years ago, and the magnets feel just as strong as it was the day it was given.   As a corollary to this question , is there a way to determine how long a permanent magnet will remain a magnet?   Addendum: Would two magnets remain a magnet for a shorter duration if they were glued in close proximity with like poles facing each other? What original Dr Who episodes are left when not watching unfinished stories? Accused of cheating on online exams Is there a way to disable the hotkeys? Last night, I migrated my MacBook's operating system from 10.10 "Yosemite" to 10.11 "El Capitan". Everything seemed to go smoothly. Happily, regarding my TeX-related work, I am able to use the GUI front ends "TeXworks" and "TeXshop" just fine since completing the OS upgrade. I use MacTeX2015, with all the latest updates installed.  Oddly, though, since completing the upgrade I can no longer run any MacTeX command-line utilities such as texdoc and kpsewhich. (Actually, as I just discovered, I can no longer run pdflatex either from a command line. However, I never do run pdflatex from a command line prompt, so that's not an urgent problem for me -- at least not for now.)  The suggestions given in the posting  do not seem to be applicable to this case, as they all seem to revolve around setting up /usr/texbin correctly -- /usr/texbin is no longer allowed under MacOSX 10.11. The document  by Herb Schulz deals in depth with adjusting settings in the GUI front ends, but it appears not to touch on what to do about /usr/texbin, which appears to be the now-missing link to TeX's command line utilities.   Pointers and solutions most welcome! Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that $$ \forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$  Then how prove $G$ is an abelian group?  Some context: Some of these commutation relations often imply that $G$ is abelian, for example , or . This looks like another example of this phenomenon, but the same techniques do not apply. Euler-Lagrange equation in curved spacetime: how to calculate $\partial \mathcal{L}/\partial(\nabla_\nu A_\mu)$? My problem is mostly with notation. When we apply the Euler-Lagrange equation to the Lagrangian scalar $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ Where $$F_{\mu\nu} = \partial_\mu A_\nu -\partial_\nu A_\mu= \nabla_\mu A_\nu -\nabla_\nu A_\mu$$ and I start with $$\frac{\partial\mathcal{L}}{\partial A_\mu}=\nabla_\nu\left(\frac{\partial \mathcal{L}}{\partial(\nabla_\nu A_\mu)}\right).$$   Now I rewrite the lagrangian as $\mathcal{L} = -\frac{1}{4}g^{\mu\alpha}g^{ \nu\beta}(\nabla_\mu A_\nu-\nabla_\nu A_\mu)(\nabla_\alpha A_\beta-\nabla_\beta A_\alpha)$ $$\mathcal{L}= -\frac{1}{4}g^{\mu\alpha}g^{\nu\beta}\left[(\nabla_\mu A_\nu)(\nabla_\alpha A_\beta)-(\nabla_\mu A_\nu)(\nabla_\beta A_\alpha)-(\nabla_\nu A_\mu)(\nabla_\alpha A_\beta)+(\nabla_\nu A_\mu)(\nabla_\beta A_\alpha) \right]$$ My problem is calculating $\frac{\partial \mathcal{L}}{\partial(\nabla_\nu A_\mu)}$, how are the rules of partial derivatives generalised to covariant ones? Also, is $\frac{\partial\mathcal{L}}{\partial A_\mu}=0$, is that correct ? Currently, the only way to link to a revision (clicking "link"), you just get  which shows what the post looked like at that revision:        Personally, I rarely use this and haven't ever found myself in a scenario (as far as I remember) where I would prefer the way it currently works to what I'm suggesting here.  Could we:   Add another link, or Change the current link   Such that it:   Displays the diff on a separate page (I'd prefer this, since the highlight and scroll thing doesn't always work properly - although perhaps it has something to do with loading images, so it might not apply here, or, if it doesn't, we could fix that instead), or          Highlights and scrolls to a diff, like what is done when you link to an answer () Consider the following two versions of a question:1   Version 1: What is the difference in meaning between "[to be] not invited" and similar negation forms?  Version 2: What is the difference in meaning among "[to be] not invited" and similar negated forms?    Personally, I think Version 1 might subtly suggest a Star Topology (see below) with the target sentence "[to be] not invited"at the center.  And Version 2 might subtly suggest a more free-form comparison like the Mesh or Fully Connected, with the target sentence simply "one among many":   Between approx equal "Star" and among approx equal "Fully Connected"?  Don't get me wrong: I do think this is quite subtle, not always applicable, and I don't think anyone would intuit such a "network topology" meaning without further explanation.   So the main question is, are both grammatically correct?  If yes, is there a semantic difference to choose one over another, perhaps as I suggested?  Or does the plurality aspect of between and among trump other semantic and stylistic issues?    1. In my question , another user provided a very helpful edit. However, the change of the word between to among prompted this question. says:     A humanoid killed by this spell rises at the start of your next turn as a zombie that is permanently under your command   If a player is hit by finger of death and dropped to 0hp, do they turn into a zombie immediately? If they subsequently fail death saves and die, do they then turn into a zombie? Why is negative times negative = positive? $ d_1,d_2\mid n\iff {\rm lcm}(d_1,d_2)\mid n\ $ [LCM Universal Property] Sudoers blacklist This is a small exercise that I just can't seem to figure out. When I see it I'll probably go 'ahhh!', but so far I haven't made any progress.  I'd like to prove that any linear functional $\phi$ on $\ell^\infty$ that is positive, i.e. for any $(x_n) \in \ell^\infty$ such that $x_n \geq 0$ for all $n \in \mathbb{N}$ we have $\phi((x_n)) \geq 0$, is a bounded linear operator and has $\|\phi \| = \phi((1,1,...))$.  Nothing I have tried worked so far, so any hints would be appreciated, thanks! Weird line in my mirrored mesh Show that $1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p$ Show that    $$1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p$$   So, I use Fermat's little theorem, that is if $p$ does not divide $a$, then $a^{p-1}$ is congruent to $1$ (mod $p$). But I'm still confused. If $1^{p-1}$ is congruent to $1$ (mod $p$), $2^{p-1}$ is congruent to $1$ (mod $p$), and $p$ does not divide $p-1$, so $(p-1)^{p-1}$ is congruent to $1$ (mod $p$). How does that equal $-1$ (mod $p$)? SharePoint 2013 REST display lookup fields? I am using SharePoint 2013 REST to show data from my list,  one of my field in announce list is AnnouncedBy which is a user packer field  or is a lookup field   how can I display it or get it using SharePoint 2013 REST?   var app = angular.module('SmartphoneOSList', ['ngSanitize']);    app.controller('ListController', function ($scope, $http) {    $http({       method: 'GET',       url: "https://myserver/_api/web/lists/getByTitle('announce')/items?$top=2&amp;$orderby=Modified desc",       headers: { "Accept": "application/json;odata=verbose" }      }).success(function (data, status, headers, config) {        $scope.items = data.d.results;       }).error(function (data, status, headers, config) {       //...     });   }); Why do I get "Exception; must be caught or declared to be thrown" when I try to compile my Java code?