我尝试使用fork（）创建一个进程树，以便每个父进程的子进程数在给定数组中，例如，如果数组是{2,1,3,0,0,0,0}，则该树将如下所示：
| a |
/ \
| b | | c |
/ / | \
| d | | e | | f | | g |
通过检查fork（）返回的值是否为0，我可以创建进程并将父进程与子进程分离。
我设法创建了一个流程树，但我设法创建的树是对称的，而不是我真正想要构建的。
我搞不清兄弟姐妹之间的路由过程，
如何分别检查每个进程应该为其创建多少子进程，并在不为其他同级进程创建的情况下为其创建？
这就是我目前所得到的：
int main() {
int nums[7] = { 2,1,3,0,0,0,0 };
int pid, pid2;
size_t len = sizeof(nums)/sizeof(int);
int childs2;
printf("\nProcess number %d has pid= %d\n", 0, getpid());
int childs = 1;
while( childs <= nums[0] ) {
pid = fork();
if (pid == 0 ) {
printf("Process number %d has pid= %d\n", childs, getpid());
printf("I am Process with pid=%d and my parent pid=%d\n", getpid(), getppid());
waitpid(getppid());
for (int i=1; i<len; i++) {
childs2 = 0;
if (childs2 < nums[i]) {
pid2 = fork();
if (pid2 == 0) {
printf("Process number %d has pid= %d\n", childs, getpid());
printf("I am Process with pid=%d and my parent pid=%d\n", getpid(), getppid());
waitpid(getppid());
break;
} else {
wait(NULL);
childs2++;
}
} else {
childs2++;
}
}
break;
} else {
wait(NULL);
childs++;
}
}
return 0;
}
我必须区分这些过程才能知道哪个过程是叶子，哪个过程是父过程。要做到这一点，我需要在每一个过程中执行不同的操作，我想不出一种方法，
我的输出是：
Process number 0 has pid= 98431
Process number 1 has pid= 98432
I am Process with pid=98432 and my parent pid=98431
Process number 1 has pid= 98433
I am Process with pid=98433 and my parent pid=98432
Process number 1 has pid= 98434
I am Process with pid=98434 and my parent pid=98432
Process number 2 has pid= 98435
I am Process with pid=98435 and my parent pid=98431
Process number 2 has pid= 98436
I am Process with pid=98436 and my parent pid=98435
Process number 2 has pid= 98437
I am Process with pid=98437 and my parent pid=98435
这棵树看起来像：
| a |
/ \
| b | | c |
/ \ / \
| d | | e || f | | g |
但我希望输出是：
Process number 0 has pid= 98431
Process number 1 has pid= 98432
I am Process with pid=98432 and my parent pid=98431
Process number 2 has pid= 98433
I am Process with pid=98433 and my parent pid=98431
Process number 3 has pid= 98434
I am Process with pid=98434 and my parent pid=98432
Process number 4 has pid= 98435
I am Process with pid=98435 and my parent pid=98433
Process number 5 has pid= 98436
I am Process with pid=98436 and my parent pid=98433
Process number 6 has pid= 98437
I am Process with pid=98437 and my parent pid=98433
所以这棵树看起来像：
| a |
/ \
| b | | c |
/ / | \
| d | | e | | f | | g |
是的。
最佳答案
我们需要做的是跟踪我们在列表中的进程以及孩子们在列表中的位置。下面的代码演示了如何执行此操作。
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#define NumberOf(a) (sizeof (a) / sizeof *(a))
/* Create children for process p.
In a child, return the number of that child process.
In the parent, return -1.
*/
static int CreateChildren(int NumberOfChildren[], int FirstChild[], int p)
{
// Create children for process p.
printf("Process %d has pid %u and parent %u.\n",
p, (unsigned) getpid(), (unsigned) getppid());
for (int i = 0; i < NumberOfChildren[p]; ++i)
{
pid_t pid = fork();
if (pid == -1)
{
perror("fork");
exit(EXIT_FAILURE);
}
if (pid == 0)
/* This is a child process, and it is child i of process p, so
its process number is FirstChild[p] + i. Return that.
*/
return p = FirstChild[p] + i;
}
// Wait for children to finish.
for (int i = 0; i < NumberOfChildren[p]; ++i)
wait(0);
// Tell caller the parent finished.
return -1;
}
int main(void)
{
int NumberOfChildren[] = { 2, 1, 3, 0, 0, 0, 0 };
size_t N = NumberOf(NumberOfChildren);
// Check the NumberOfChildren array for consistency.
{
int sum = 0;
for (size_t n = 0; n < N; ++n)
{
if (NumberOfChildren[n] < 0)
{
fprintf(stderr,
"Error, number of children cannot be negative but is %d.\n",
NumberOfChildren[n]);
exit(EXIT_FAILURE);
}
sum += NumberOfChildren[n];
}
if (sum != N-1)
{
fprintf(stderr,
"Error, the numbers of children sum to %d desecendants "
"of the root, but array has %zu elements after the root "
"element.\n",
sum, N-1);
exit(EXIT_FAILURE);
}
}
/* Compile information about the children -- set FirstChild[n] to the
index of the element in NumberOfChildren that is for the first child
of process n.
*/
int FirstChild[N];
{
int NextChild = 1;
for (int n = 0; n < N; ++n)
{
FirstChild[n] = NextChild;
NextChild += NumberOfChildren[n];
}
}
// This is the root process. Set p to its index.
int p = 0;
/* Create children for process p. When a child is created, it will
return its process number, and we will loop to create children for it.
*/
while (p >= 0)
p = CreateChildren(NumberOfChildren, FirstChild, p);
}
样本输出：
进程0具有PID 2648和父进程2641。
进程1具有PID 2649和父进程2648。
进程2具有pid 2650和父进程2648。
进程3具有pid 2651和父进程2649。
进程4具有pid 2652和父进程2650。
进程5具有pid 2653和父进程2650。
进程6具有pid 2654和父进程2650。
关于c - 有办法更好地路由流程吗？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55500369/