GCC 4.4 will give you a warning. In the later versions of GCC, this feature (-Wunreachable-code) has been removed. See Re: gcc -Wunreachable-code option The -Wunreachable-code has been removed, because it was unstable: it relied on the optimizer, and so different versions of gcc would warn about different code. The compiler still accepts and ignores … Read more
Yes, it does normally translate into an implicit if statement with an internal boolean flag. So, in the most basic implementation your declaration normally translates into something like void go( int x ) { static int j; static bool j_initialized; if (!j_initialized) { j = x; j_initialized = true; } … } On top of … Read more
Your description of a static variable is more fitting to that found in C. The concept of a static variable in Object Oriented terms is conceptually different. I’m drawing from Java experience here. Static methods and fields are useful when they conceptually don’t belong to an instance of something. Consider a Math class that contains … 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.
In the context of class attributes, static has a different meaning. If you have a field like: private static int sharedAttribute; then, each and every instance of the class will share the same variable, so that if you change it in one instance, the change will reflect in all instances, created either before or after … 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
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
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’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
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