How to change local static variable value from outside function

Question

#include 

void func() {
   static int x = 0;  // x is initialized only once across three calls of
                      //     func()
   printf(\"%d\\n\"         


        

Answer 1

You can't. It's a local static variable which is invisible outside func().

You can either use:

1. Global static (not recommended generally).
2. Pass it through reference/pointer parameter, or return by reference.

Answer 2

Do you want the function always to reset the counter after three invocations? or do you want to the caller to reset the count at arbitrary times?

If it is the former, you can do it like this:

void func() {
  static int x = 0;
  printf("%d\n", x);
  x = (x + 1) % 3;
}

If it is the latter, using a local static variable is probably a bad choice; you could instead use the following design:

class Func
{
  int x;
  // disable copying

public:
  Func() : x(0) {}

  void operator() {
    printf("%d\n", x);
    x = x + 1;
  }

  void reset() {
    x = 0;
  }
};

Func func;

You should make it either non-copyable to avoid two objects increasing two different counters (or make it a singleton), or make the counter a static member, so that every object increments the same counter. Now you use it like this:

int main(int argc, char * const argv[]) {
  func(); // prints 0
  func(); // prints 1
  func(); // prints 2

  func.reset();
  return 0;
}

Answer 3

You can have func() assign the address of the variable to a pointer that's visible from outside func().

Or you can have a special parameter you can pass to func() to tell it to reset the variable.

But really, the whole point of declaring x inside func() is to make it visible only within the body of func(). If you want to be able to modify x, define it somewhere else. Since you're writing C++, it probably makes sense for x to be a static member of some class.