基本的c问题-在c中声明静态变量时会发生什么？

Question

I'm trying to learn how static variable work in c for when they are defined in a given function. For example, when I write the following:

#include <stdio.h>
void inc() {
  static int c = 0;
  c++;
  printf("%d\n", c);
}

int main(void) {
  inc();
  inc();
  inc();
  return 0;
}

The expected output is obviously:

1
2
3

On the first call of the function, the static variable c is defined and given the value of 0, which makes perfect sense. It is the incremented and printed. However, on the second call to inc() why is it that the integer c is maintained and not set to zero, even though the code literally says static int c = 0; . What mechanism in the compiler stops c from having it's value set to zero like during the first call?

Answer 1

Quoting C11 , chapter §6.2.4, Storage durations of objects ( emphasis mine )

An object whose identifier is declared without the storage-class specifier _Thread_local , and either with external or internal linkage or with the storage-class specifier static , has static storage duration . Its lifetime is the entire execution of the program and its stored value is initialized only once, prior to program startup.

So, the initialization inside the function call does not take place on every call to the function. It only happens once, before the execution of main() starts. The variable retains the last stored value through the program execution, ie, the value is retained between the repeated calls to the function.

Answer 2

它的生命周期是程序的整个执行过程，并且其值在程序启动之前被初始化并且只有一次。