fork()和更改局部变量? [英] fork() and changing local variables?

问题描述

我试图理解fork()的概念，有一件事我似乎无法理解.

I am trying to understand the fork() concept and there's is one thing I can't seem to understand.

在以下代码中-为什么即使子进程将其更改为5，父级仍会打印i = 0?

In the following code - why does the parent still print i=0 even when child process changes it to 5?

wait(NULL)阻止父进程，直到子进程先完成.

The wait(NULL) blocks parent process until child finishes first.

int main(int argc, char *argv[]) {
  int i = 0;
  if (fork() == 0) {
    i = 5;
  } else {
    wait(NULL);
    printf("i = %d\n", i);
  }
  return 0;
}

有人可以解释为什么我的假设不正确吗?

Can somebody explain why my assumption is incorrect?

推荐答案

进程之间不共享变量.调用fork之后，有两个完全独立的过程. fork在子级中返回0，其中局部变量设置为5.在父级中，fork返回子级的进程ID，i的值不变.在调用fork之前，它仍然具有设置为0的值.就像您有两个程序单独运行一样，行为相同:

Variables are not shared between processes. After the call to fork, there are two completely separate processes. fork returns 0 in the child, where the local variable is set to 5. In the parent, where fork returns the process ID of the child, the value of i is not changed; it still has the value 0 set before fork was called. It's the same behavior as if you had two programs run separately:

int main(int args, char *argv[]) {
    int i=0;
    printf("i = %d\n", i);
    return 0;
}

和

int main(int argc, char *argv[]) {
  int i = 0;
  i = 5;
  return 0;
}

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