关于Linux -Copy-on-write机制的疑惑

kunis

代码如下：
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>

volatile static int g_value = 0xFF;
static int *pbuf =NULL;
void child_task()
{
    sleep(10);
    while(1){
       printf("child task[%d] value:%d address:0x%p,*pbuf = %d\n",getpid(),g_value,&g_value,*pbuf);
       sleep(2);
    }
    exit(0);
}
int main(int argc,char *argv[])
{
  pid_t pid = -1;
  pbuf = (int *)malloc(4);
  *pbuf = 0;

  pid = fork();
  if(pid == 0){

      child_task();
  }
  while(1){

      printf("parent task[%d] value:%d address:0x%p,*pbuf = %d\n",getpid(),g_value,&g_value,*pbuf);
      g_value++;
      *pbuf = *pbuf +1;
      sleep(4);
  }
  exit(0);
  return 0;

}

parent task[2242] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:256 address:0x0x804a024,*pbuf = 1
parent task[2242] value:257 address:0x0x804a024,*pbuf = 2
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:258 address:0x0x804a024,*pbuf = 3
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:259 address:0x0x804a024,*pbuf = 4
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:260 address:0x0x804a024,*pbuf = 5
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:261 address:0x0x804a024,*pbuf = 6
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:262 address:0x0x804a024,*pbuf = 7
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
parent task[2242] value:263 address:0x0x804a024,*pbuf = 8
child task[2243] value:255 address:0x0x804a024,*pbuf = 0
child task[2243] value:255 address:0x0x804a024,*pbuf = 0

编译运行的环境为Ubuntu 10.04.4 LTS／Linux 2.6.32-38-generic
从log来看 子进程继承了父进程变量以及heap。
根据网上COW的解释，子进程再去写内存的时候才会复制一份父进程的物理内存，如果不写，则子进程的page table表用的是父进程。

现在很困惑，既然以上程序 子进程没有写g_value，按理讲子进程g_value的值应该跟父进程指向同一个物理地址。

或者COW的理解为  只要是父进程或子进程的任何一个去写变量，都会触发copy page table？？？？

timespace

打印的地址是虚拟内存地址，不是物理地址。