C++二进制fork UNIX中的特定数字

我正在学习fork如何在类Unix操作系统中工作。我想创建9个子进程，每个父进程只能创建2个子进程。所以这是一个二进制fork算法。
然而，我不能设法限制在第三次递归调用中创建的孩子的正确数量。
下面是创建子节点的流程。

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;

}

字符串
下面是输出：

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

型
第一个调用生成2个子元素，第二个调用生成4个子元素，第三个调用生成4个子元素，但应该是3个。

你希望每个进程最多执行两次fork()，所以你不需要任何循环、递归之类的东西。
给定一个命令行参数9，下面的程序will print 10倍Terminating.和9倍Waited for: <pid>，这似乎是你的绘图指定的。

#include <errno.h>
#include <sys/wait.h>
#include <unistd.h>

#include <cstdint>
#include <cstdlib>
#include <iostream>
#include <sstream>

namespace {
struct Process {
  Process() : pid{fork()} { 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;
  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;
  if (secondFork.is_child()) {
    firstFork.disarm();
    if (--remaining) goto fork_again;
  }
}

字符串
只要有两个无约束的fork()，它就相当于Bash的强大功能：

_()(_&_&);_

型
为了驯服进程数量的指数增长，我们限制了一个进程在fork() ing树上运行fork()的次数，如果我们希望fork() s的总数保持不变，这个数量需要指数下降。
剩下的就是正确处理细节了。第一个子树被授予fork() s的remaining / 2 + remaining % 2，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（grand）父进程之外的所有进程也必须计数 * 自身 *，这就是为什么它们总是以--remaining开始。
由于the goto的原因，在这段代码中使用RAII在可读性方面是有争议的。我几乎可以肯定有一种方法可以实现它而不需要向后跳转的goto语句，但我也懒得去想这个问题。

C++二进制fork UNIX中的特定数字

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