I have a requirement where I need to Deserialize and Create few objects of a class(Table) which doesn't have a Default Constructor.  Snippet of my code   else if (reader.Name == "Tables") {     reader.ReadStartElement();     tables = SerializationHelper&lt;Table&gt;.DeserializeList(reader); }   The definition of DeserializeList in SerializationHelper is as below:  public static List&lt;T&gt; DeserializeList(XmlReader reader) {     XmlSerializer ser = new XmlSerializer(typeof(T));     List&lt;T&gt; returnval = new List&lt;T&gt;();     while (reader.NodeType != XmlNodeType.EndElement)     {         T result = (T)ser.Deserialize(reader);         returnval.Add(result);     }     return returnval; }   This is an existing working code and with recent changes we had to add a mandate parameter to all the Constructors in the Class  The Table class here doesn't have any Parameter-less Constructor now.  I wanted to find out if I can by anyway pass-on at-least one parameter when DeSerializing the Table objects.  I have already Read the following but they use JSON.net which in my case is not an option to use.
I'm writing code to do Xml serialization. With below function.  public static string SerializeToXml(object obj) {     XmlSerializer serializer = new XmlSerializer(obj.GetType());     using (StringWriter writer = new StringWriter())     {         serializer.Serialize(writer, obj);         return writer.ToString();     } }   If the argument is a instance of class without parameterless constructor, it will throw a exception.     Unhandled Exception:   System.InvalidOperationException:   CSharpConsole.Foo cannot be serialized   because it does not have a   parameterless constructor.    at   System.Xml.Serialization.TypeDesc.CheckSupported()   at   System.Xml.Serialization.TypeScope.GetTypeDesc(Type   type, MemberInfo sourc e, Boolean   directReference, Boolean throwOnError)   at   System.Xml.Serialization.ModelScope.GetTypeModel(Type   type, Boolean direct Reference)    at   System.Xml.Serialization.XmlReflectionImporter.ImportTypeMapping(Type   type , XmlRootAttribute root, String   defaultNamespace)    at   System.Xml.Serialization.XmlSerializer..ctor(Type   type, String defaultName space)    at   System.Xml.Serialization.XmlSerializer..ctor(Type   type)   Why must there be a parameterless constructor in order to allow xml serialization to succeed?  EDIT: thanks for cfeduke's answer. The parameterless constructor can be private or internal.
In my preamble, you'll find:  \renewcommand{\baselinestretch}{1} \renewcommand{\familydefault}{\sfdefault} \usepackage[sfmath]{kpfonts} \usepackage[mathrmOrig]{sfmath} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{animate} \usepackage{anysize} \usepackage{arcs} \usepackage{array} \usepackage[scale=0]{background} \usepackage{blindtext} \usepackage{calc} \usepackage{cancel} \usepackage{chngcntr} \usepackage{CJK} \usepackage{CJKspace} \usepackage{CJKpunct} \usepackage{colortbl} \usepackage{enumitem} \usepackage{epstopdf} \usepackage{etoolbox} \usepackage{fancybox} \usepackage{fancyhdr} \usepackage{float} \usepackage{framed} \usepackage{graphicx} \usepackage{hyphenat} \usepackage{ifthen} \usepackage{latexsym} \usepackage{lmodern} \usepackage{multicol} \usepackage{multirow} \usepackage{mwe} \usepackage{pgffor} \usepackage{rotating} \usepackage{tabularx} \usepackage{textcomp} \usepackage{tgheros,tgtermes,tgcursor} \usepackage{tikz} \usepackage{times} \usepackage{tipa} \usepackage{varwidth} \usepackage{wasysym} \usepackage{wboiboites} \usepackage[abs]{overpic} \usepackage[breaklinks,colorlinks=true,linkcolor=red,citecolor=red, urlcolor=blue]{hyperref} \usepackage[framemethod=TikZ]{mdframed} \usepackage[breakable, theorems, skins]{tcolorbox,empheq} \usepackage[default]{gillius} \usepackage[document]{ragged2e} \usepackage[latin1]{inputenc} \usepackage[spanish]{babel} \usepackage[T1]{fontenc} \usepackage[total={18.59cm,26.5cm},left=1.5cm,top=2.5cm]{geometry} \usetikzlibrary{matrix,arrows, positioning,shadows,shadings,backgrounds,calc, shapes, tikzmark} \usetikzlibrary{shadows.blur} \usetikzlibrary{shapes.symbols} \usetikzlibrary{patterns,fadings} \tcbuselibrary{skins,breakable,listings,theorems}    I'm trying to draw an arrow for X axis and Y axis.   \begin{tikzpicture} \draw[yellow!10, fill=blue!5](-1, -1) rectangle (7, 5); \draw[dotted] (-0.75,-0.75) grid (6.75,4.75); \node[below] at (-0.25,0){$O$}; \draw[-&gt;,black, thick] (-0.75,0)--(6.75,0); \end{tikzpicture}   But I get this error:    Additionally, if I remove -> from the options of \draw the tikz picture compiles normally. "Timed Out" error when trying to join server How to remove a filter that is an anonymous object? Placing however at the start of a sentence vs placing it in the middle of a sentence? Basically, what's the difference between this:     However, it didn't mean that I didn't have the potential to become a   rock star.   and this?     It didn't mean, however, that I didn't have the potential to become a   rock star.   Or does it mean exactly the same thing? Can I download and play games from a friend's PSN account on a PS4? Using a wi-fi and an ethernet connection at the same time thus having more bandwidth? How to get a variable name as a string in PHP? Say i have this PHP code:  $FooBar = "a string";   i then need a function like this:  print_var_name($FooBar);   which prints:  FooBar   Any Ideas how to achieve this? Is this even possible in PHP? Why didn't all the Jedi at the Jedi Temple sense Anakin and his legion of Clones’ thoughts? If you go back to Attack of the Clones, Anakin and Obi-Wan sensed that poisonous alien worms were about to kill Padme. Back in Revenge of the Sith, the Jedi could take on the legion or if there were too many, they could at least escape. Perturbation property of Orthonormal basis for Hilbert space Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function. Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.  I know that for $f$ is differentiable in $(a,b)$ and that it has one-sided derivative $f_+' (a)≠f_-' (b)$ at the endpoints. If $C$ is a real number between $f_+' (a)$  and $f_-' (b)$, then there exists $c∈(a,b)$ such that $f' (c)=C $. How can I use this to prove the above? How do I change my update server in Ubuntu 18.04? Double integral with transformation; possible error in limits I'm asked to show that the following integral is true with the transformation $u = x+y$ and $y=uv$:  $$\int_0^1 \int_0^{1-x} e^{y/(x+y)} dx dy = \frac{e-1}2 $$  I found the determinant of the jacobian of the transformation is $-u$, and the exponent is just $v$ so the integral becomes $$\iint_R |-u|e^v du dv = \frac {u^2}2 e^v|_R$$ $$=\frac{(x+y)^2}2e^{y/(x+y)}|_{x=0}^{1-x}\;|_{y=0}^1$$ And plugging it all in doesn't work (AFAICT). For one, the x's don't all cancel.  I've looked at it 100 times. It's definitely written dxdy not dydx even though there is definitely an x in the integral. Maybe this is a typo and why my calculation isn't working out? I tried reversing the order to dydx and it didn't work. I tried changing the limit of integration to $1-y$ and it didn't work out.  I'm not even sure how to figure out the domain of integration in the uv-plane. Flipping things around yields y=uv and x=u-uv. But if my domain is bounded by x=0 and y=0 and y=1 (ignoring the last limit for now), then I have uv=o=u and uv=1. !?!  Am I just making an algebra mistake plugging everything in? Is there a way to handle a variable in it's own integral that I'm unaware of? $\LaTeX$ and Mathematica Derivative of inverse function at a point. Let $f\left(x\right)=x^{13}+3x^{9}+2x^{5}+x^{2}+x+2$ find $\left(f^{-1}\right)'\left(1\right)$. Can somebody show me the method? I am a little confused. Update  appear to have been updated recently (as of writing this the last update appears to have taken place on 3rd of June) to include the following statement:  The -d switch is necessary to upgrade from Ubuntu 18.04 LTS as upgrades have not yet been enabled and will only be enabled after the first point release of 20.04 LTS.  Also note that .  TLDR (as of 15 Aug 2020):   is the canonical tracking document for the first Focal Fossa point-release (20.04.1). It’s a live document. The Ubuntu release team will be updating it as we work on releasing 20.04.1. […] Upgrades from 18.04 to 20.04.1 are still disabled as we are working through a few upgrade blockers   Edit The answers (so far) do not answer the question. The question is pretty simple. Ubuntu.com points to a &quot;FinalRelease&quot;, which turns out to be a &quot;Development Release&quot;, see evidence below. I'm not asking how and when to upgrade. I'm pointing out what seems to me to be an apparent inconcistency (in semantics, linguistics, logic) in the way Ubuntu names or labels its releases. From an answer to the question I'd except someone to pick up and acknowledge this inconsistency. An answer could start with a &quot;Yes, ...&quot; or a &quot;No, ...&quot;, for starters. (Moreover, a good answer could distinguish between the release being in dev stage and/or the upgrade process (e.g. from LTS to LTS) being in dev stage, if appropriate.) This question, as I see it, is not a duplicate. Futher, I think it is a question which deserves a non-patronizing answer. Thanks.   indicates that today (23rd April 2020) the new Ubuntu 20.04 LTS is to be released.  See &quot;condensed screenshot&quot; here:  Indeed, it seems that that's what happened, see  and :  However, when following the instructions to upgrade (from 18.04 LTS to 20.04 LTS) found  and elsewhere (edit: actually , too!), even prior to the July point release, I'm a bit confused to see that the instruction is to use update-manager -d which also points to the &quot;Development Release&quot; (see screenshot below) rather than the &quot;FinalRelease&quot; (see screenshot and link above). Is the &quot;Final Release&quot; (which I'm after) the/a &quot;Development Release&quot;? Why isn't it called something like &quot;Production Release&quot; or &quot;Final Release&quot;? When will it no longer be referred to as &quot;Development Release&quot;? In July? I'm confused :( Random test data: How to do it right? When I look in sys.sql_logins, I see a column called is_policy_checked. Can I trust that my password policy has been checked for all of the logins where this column value is 1? Access 2007 Work in a separate file using ACCDE I'd like to declare some new operators using \DeclareMathOperator and some of the names are already taken by other commands (that I do not want to use). Is there an equivalent to the \renewcommand command for \DeclareMathOperator or a different way to overwrite the commands using \DeclareMathOperator? So, @AndrewO mentioned recently that he has had 'encounters' with people wondering why D-Wave has a lot more qubits than IBM. Of course, this comparison is faulty, since the IBM and D-Wave's machine may both exploit quantum effects to a certain degree, IBM's machine matches the thing the TCS people call a 'Quantum computer' a bit more than D-Waves's alleged quantum annealer.   How do you explain to a novice why IBM is still reaching important milesstones, even though the D-Wave has a lot more 'qubits'. I understand that an easy answer is, 'well, you're comparing apples and pears', but that is merely relying on your possible authority and simply doesn't explain anything!   How can you explain that those devices are different, how can you dispell the myth that the number of qubits is not the only metric to judge quantum devices? (preferably to a layman, but assuming basic (undergrad?) physics knowledge is ok, if needed)