Question

I noticed that if you initialize a static variable in C++ in code, the initialization only runs the first time you run the function.

That is cool, but how is that implemented? Does it translate to some kind of twisted if statement? (if given a value, then ..)

void go( int x )
{
    static int j = x ;
    cout << ++j << endl ; // see 6, 7, 8
} 

int main()
{
    go( 5 ) ;
    go( 5 ) ;
    go( 5 ) ; 
}

Answer 1

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 that, if your static object has a non-trivial destructor, the language has to obey another rule: such static objects have to be destructed in the reverse order of their construction. Since the construction order is only known at run-time, the destruction order becomes defined at run-time as well. So, every time you construct a local static object with non-trivial destructor, the program has to register it in some kind of linear container, which it will later use to destruct these objects in proper order.

Needless to say, the actual details depend on implementation.

---

It is worth adding that when it comes to static objects of “primitive” types (like int in your example) initialized with compile-time constants, the compiler is free to initialize that object at startup. You will never notice the difference. However, if you take a more complicated example with a “non-primitive” object

void go( int x ) {
  static std::string s = "Hello World!";
  ...

then the above approach with if is what you should expect to find in the generated code even when the object is initialized with a compile-time constant.

In your case the initializer is not known at compile time, which means that the compiler has to delay the initialization and use that implicit if.