A static variable shares the value of it among all instances of the class. Example without declaring it static: public class Variable { public int i = 5; public void test() { i = i + 5; Console.WriteLine(i); } } public class Exercise { static void Main() { Variable var = new Variable(); var.test(); Variable … Read more
Fundamentally this is because static members must be defined in exactly one translation unit, in order to not violate the One-Definition Rule. If the language were to allow something like: struct Gizmo { static string name = “Foo”; }; then name would be defined in each translation unit that #includes this header file. C++ does … Read more
You’re basically referring to the section ยง7.3.1.1/2 from the C++03 Standard, The use of the static keyword is deprecated when declaring objects in a namespace scope; the unnamed-namespace provides a superior alternative. Note that this paragraph was already removed in C++11. static functions are per standard no longer deprecated! Nonetheless, unnamed namespace‘s are superior to … Read more
Instead of bar use self.bar or Foo.bar. Assigning to Foo.bar will create a static variable, and assigning to self.bar will create an instance variable.
You have answered your own question. Static initialization order is undefined, and the most elegant way around it (while still doing static initialization i.e. not refactoring it away completely) is to wrap the initialization in a function. Read the C++ FAQ items starting from https://isocpp.org/wiki/faq/ctors#static-init-order