From the “New Features in C# 4” word doc: Dynamic lookup will not be able to find extension methods. Whether extension methods apply or not depends on the static context of the call (i.e. which using clauses occur), and this context information is not currently kept as part of the payload.
You can try this: dynamic task = JObject.Parse(rawJson); Documentation: Querying JSON with dynamic
To check if the assembly is dynamic: if (assembly.ManifestModule is System.Reflection.Emit.ModuleBuilder) This took me a while to figure out, so here it is asked and answered. Update: In .NET 4.0, there is now a property: if (assembly.IsDynamic)
You need to add a reference to Microsoft.CSharp.dll. This provides the required types for using dynamic in C#. Also, you will likely need to upgrade to IronPython 2.7[.3] or later, as there are some incompatibilities with old releases and the newer .NET frameworks.
Great question. You basically have a complex trade-off between: Speed of a complete collision detection phase Overhead of updating / maintaining the data structure as objects move around The bad news is that there is no “perfect” answer because of this – it will genuinely depend on lots of different factors unique to your situation. … Read more
Try this LinearLayout.LayoutParams layoutParams=new LinearLayout.LayoutParams(width, height); layoutParams.gravity=Gravity.CENTER; ImageIcons[i].setLayoutParams(layoutParams);
You seem to be mixing three completely different, orthogonal things: static vs. dynamic typing manifest vs. implicit typing named vs. anonymous types Those three aspects are completely independent, they have nothing whatsoever to do with each other. Static vs. dynamic typing refers to when the type checking takes place: dynamic typing takes place at runtime, … Read more
I don’t think there is black and white here – there is a whole spectrum between dynamic and static. Let’s take two extreme examples for each side of the spectrum, and see where that takes us. Haskell is an extreme in the static direction. It has a powerful type system that is checked at compile … Read more
That should work, yes. You can also do: $f = new $class($arg1,$arg2);