我正在学习 fork 如何与类 Unix 操作系统一起工作。 我想创建 9 个子进程,每个父进程只能创建 2 个子进程。 所以这是一个二分叉算法。
但是我无法限制在第三次递归调用中创建的正确子级数量。
这是我现在拥有的代码。
#include <iostream.h>
#include <sys/wait.h>
#include <unistd.h>
#include <stdlib.h>
using namespace std;
static int CHILD_SIZE = 9;
static int LEVEL = 1;
static int MAX_LEVEL = 3;
void createChildren(int remaining, int level) {
int CURRENT_REMAINING = remaining;
if(CURRENT_REMAINING <= 0 || level > MAX_LEVEL){
return;
}
pid_t firstChild,secondChild;
// Fork the first child
firstChild = fork();
if(firstChild == -1){
cerr << "Fork failed!" << endl;
}
else if (firstChild == 0 && CURRENT_REMAINING > 1)
{
// first child
cout << "First child process (PID: " << getpid() << ") with parent process (PID: " << getppid() << ") at level: " << level << endl;
// calculate how many child needs after this iteration
CURRENT_REMAINING -= (2 * level);
if(CURRENT_REMAINING < 0){
CURRENT_REMAINING = 0;
}
}
else if (firstChild > 0 && CURRENT_REMAINING > 1)
{
// Fork the second child
secondChild = fork();
if(firstChild == -1){
cerr << "Fork failed!" << endl;
}
else if (secondChild == 0) {
// second child
waitpid(firstChild, NULL, 0);
cout << "Second child process (PID: " << getpid() << ") with parent process (PID: " << getppid() << ") at level: " << level << endl;
// calculate how many child needs after this iteration
CURRENT_REMAINING -= (2 * level);
if(CURRENT_REMAINING < 0){
CURRENT_REMAINING = 0;
}
} else if (secondChild > 0){
// Wait for both child processes to complete
waitpid(firstChild, NULL, 0);
waitpid(secondChild, NULL, 0);
}
}
//---------- recursion call-----------//
// upper child recursion call
if(firstChild == 0){
cout << "current remaining in first child is: " << CURRENT_REMAINING << " at level " << level << endl;
createChildren(CURRENT_REMAINING, level + 1);
}
// lower child recursion call
if(secondChild == 0){
cout << "current remaining in second child is: " << CURRENT_REMAINING << " at level " << level << endl;
if(CURRENT_REMAINING > 3){
//cout << "Create two more processes" << endl;
createChildren(CURRENT_REMAINING, level + 1);
}
}
}
int main() {
std::cout << "Parent process " << getpid() << std::endl;
createChildren(CHILD_SIZE, LEVEL);
return 0;
}
这是输出:
Parent process 8571
First child process (PID: 8573) with parent process (PID: 8571) at level: 1
current remaining in first child is: 7 at level 1
Second child process (PID: 8574) with parent process (PID: 8571) at level: 1
current remaining in second child is: 7 at level 1
First child process (PID: 8575) with parent process (PID: 8573) at level: 2
current remaining in first child is: 3 at level 2
First child process (PID: 8576) with parent process (PID: 8574) at level: 2
current remaining in first child is: 3 at level 2
Second child process (PID: 8577) with parent process (PID: 8573) at level: 2
current remaining in second child is: 3 at level 2
Second child process (PID: 8578) with parent process (PID: 8574) at level: 2
current remaining in second child is: 3 at level 2
First child process (PID: 8579) with parent process (PID: 8575) at level: 3
current remaining in first child is: 0 at level 3
Second child process (PID: 8581) with parent process (PID: 8575) at level: 3
current remaining in second child is: 0 at level 3
First child process (PID: 8580) with parent process (PID: 8576) at level: 3
current remaining in first child is: 0 at level 3
Second child process (PID: 8582) with parent process (PID: 8576) at level: 3
current remaining in second child is: 0 at level 3
第一个电话就生了两个孩子 第二个电话产生了 4 个孩子 第三次调用得到 4,但应该是 3。
您希望每个进程最多
fork()
两次,因此您不应该需要任何循环、递归或类似的东西。
给定命令行参数
9
,以下程序将打印 10次Terminating.
和9次Waited for: <pid>
,这似乎是您的图片所指定的。
#include <sys/wait.h>
#include <unistd.h>
#include <cstdint>
#include <cstdlib>
#include <iostream>
#include <sstream>
namespace {
struct Process {
Process(pid_t pid) : pid{pid} { if (pid == -1) throw -1; }
bool is_child() const { return pid == 0; }
void disarm() { pid = 0; }
~Process() { if (pid)
std::cout << (std::stringstream{}
<< "Waited for: " << waitpid(pid, nullptr, 0) << '\n').str(); }
private:
pid_t pid;
};
struct Blah { ~Blah() { std::cout << "Terminating.\n"; } };
} // namespace
int main(int argc, const char *const argv[]) {
Blah blah;
if (argc != 2) return EXIT_FAILURE;
size_t remaining;
if (!(std::stringstream{argv[1]} >> remaining)) return EXIT_FAILURE;
fork_again:
const size_t old_remaining{remaining};
remaining = old_remaining / 2 + old_remaining % 2;
if (!remaining) return EXIT_SUCCESS;
Process firstFork{fork()};
if (firstFork.is_child()) {
if (--remaining) goto fork_again; else return EXIT_SUCCESS;
}
remaining = old_remaining - remaining;
if (!remaining) return EXIT_SUCCESS;
const Process secondFork{fork()};
if (secondFork.is_child()) {
firstFork.disarm();
if (--remaining) goto fork_again;
}
}
看待它的一种方式是,只需两个不受约束的
fork()
,它就相当于 Bash 中的这种令人敬畏的功能:
_()(_&_&);_
为了抑制进程数量的指数增长,当我们沿着
fork()
树向上移动时,我们限制进程可以 fork()
的次数,并且该数字呈指数下降。剩下的就是让所有细节(在某种程度上)正确。第一个子树被授予 remaining / 2 + remaining % 2
中的 fork()
,第二个子树获得其余的。这可以正确处理奇数和偶数情况。此外,除了原始(祖)父进程之外的所有进程也必须计算它们自己,这就是为什么它们总是以--remaining
开始。
goto
,在这段代码中使用 RAII 在可读性方面存在一些争议。我几乎确信有一种方法可以实现这一点,而无需向后跳转 goto
语句,但我也懒得弄清楚这一点。