Having this function
public ParameterMethodParameterBuilder withParameter() {
MethodParameter parameter = new MethodParameter();
return withParameter(parameter).new ParameterMethodParameterBuilder(parameter);
}
What is the mean of the experession below
withParameter(parameter).new ParameterMethodParameterBuilder(parameter)
The syntax obj.new Inner() creates and returns an instance of the inner class(*) Inner that is linked to the instance obj of the encapsulating class.
When an inner class is declared, you need an instance of the encapsulating class to instantiate the inner class. The syntax you are confronted to is exactly for this purpose.
Here is the simplest example for this:
public class MainClass {
public class InnerClass {
}
}
You would instantiate InnerClass this way:
MainClass mc = new MainClass();
mc.new InnerClass();
(*) inner class = non-static nested class
ParameterMethodParameterBuilder is an inner class of whatever returning type of withParameter(MethodParameter parameter) method.
You care creating a new object of that inner class ParameterMethodParameterBuilder which having outer reference returned by withParameter(parameter) method
It creates a new instance of a nested class:
public class MethodParameter() {
public class ParameterMethodParameterBuilder(/* ... */) {
//...
}
//...
}
ParameterMethodParameterBuilder is declared static? I suppose exactly that it isn't.
static.
withParameter(parameter) method returns MethodParameter instance and you are trying to create
object of ParameterMethodParameterBuilder static inner class,
so syntax goes like this
MethodParameter.new ParameterMethodParameterBuilder (parameter) to create your inner static class object
let me know for any issues.
withParameter(MethodParameter parameter)method