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IntelliJ gives Fatal Error: Unable to find package java.lang in classpath or bootclasspath
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Somehow the JDK language jars were removed from the classpath. All I had to do was reapply the JSDK home path.
Goto:
File -> Project Structure -> Platform Settings -> SDKs
Re-apply the JSDK home path.
Doing this added about 15 jars to the classpath. Apparently these are impo...
Python 3.x rounding behavior
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Python 3's way (called "round half to even" or "banker's rounding") is considered the standard rounding method these days, though some language implementations aren't on the bus yet.
The simple "always round 0.5 up" technique results in a sli...
@ variables in Ruby on Rails
...cope (current block)
@title is an instance variable - and is available to all methods within the class.
You can read more here:
http://strugglingwithruby.blogspot.dk/2010/03/variables.html
In Ruby on Rails - declaring your variables in your controller as instance variables (@title) makes them ava...
Can a C# class inherit attributes from its interface?
...h is searched for the attributes.</param>
/// <returns>Returns all attributes.</returns>
public static T[] GetCustomAttributes<T>( this Type type ) where T : Attribute
{
return GetCustomAttributes( type, typeof( T ), false ).Select( arg => (T)arg ).ToArray();
}
/// <...
.NET 4.0 has a new GAC, why?
... Global Assembly Cache (GAC), you will have to manage each of them individually.
In .NET Framework 4.0, the GAC went through a few changes. The GAC was split into two, one for each CLR.
The CLR version used for both .NET Framework 2.0 and .NET Framework 3.5 is CLR 2.0. There was no need in the prev...
How to use gradle zip in local system without downloading when using gradle-wrapper
...es a main point of the wrapper, which is that Gradle is downloaded and installed automatically for all build users. If you prefer to use a local installation, put it on your PATH and invoke gradle rather than gradlew.
– Peter Niederwieser
Apr 7 '14 at 6:59
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Setting default value for TypeScript object passed as argument
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Actually, there appears to now be a simple way. The following code works in TypeScript 1.5:
function sayName({ first, last = 'Smith' }: {first: string; last?: string }): void {
const name = first + ' ' + last;
console.log(na...
Standard deviation of a list
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The functions are actually called stdev and pstdev, not using std for standard as one would expect. I couldn't edit the post as edits need to modify at least 6 chars...
– mknaf
Jan 6 '17 at 16:00
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Declaring abstract method in TypeScript
...because it is abstract. This is part of TypeScript 1.6, which is now officially live.
abstract class Animal {
constructor(protected name: string) { }
abstract makeSound(input : string) : string;
move(meters) {
alert(this.name + " moved " + meters + "m.");
}
}
class Snake ...
Contains method for a slice
...re with map[string] struct{}. map[string] struct{} seems like a hack especially initializing an empty struct struct {}{}
– vadasambar
Sep 12 '19 at 12:17
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