C static variables initialization-CodePudding

I learned some C and came across an explanation for static variables. They showed this code:

#include<stdio.h>
int fun()
{
  static int count = 0;
  count  ;
  return count;
}
  
int main()
{
  printf("%d ", fun());
  printf("%d ", fun());
  return 0;
}

I can't understand why calling the function twice is fine, because the line

static int count = 0;

actually runs twice... I can't understand how is that possible... Can you actually declare it twice or does the compiler just ignore it the second time?

CodePudding user response：

This (statics/globals) is where an initializing definition is really different from an uninitialized definition followed by an assignment. Historically, the former even used to have different syntax (int count /*no '=' here*/ 0;).

When you do:

int fun() {
  static int count = 0;
  //...
}

then except for the different scopes (but not lifetimes) of count, it's equivalent to:

static int count = 0; //wider scope, same lifetime
int fun() {
  
  //...
}

In both cases, the static variable becomes initialized at load time, typically en-masse with other statics and globals in the executable.

CodePudding user response：

static variables are initialized on program startup, not every time the function is called.

CodePudding user response：

... because the line static int count = 0; actually runs twice.

No. Just once, right before main() is called.