我尝试使用fork（）创建进程树，以使每个父级的子代数在给定的数组中，例如，如果数组为{2,1,3,0,0,0,0}树看起来像这样：
| a |
/ \
| b | | c |
/ / | \
| d | | e | | f | | g |
我能够通过检查fork（）返回的值是否为0来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，但是我设法创建的树是对称的，而不是我真正想要构建的树。 我无法弄清楚兄弟姐妹之间路由过程的一部分，
如何为每个进程分别检查应该为多少个子进程创建一个子进程，而又不为其他同级进程创建子进程呢？
这是我到目前为止所得到的：
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;
}
我必须区分进程，才能知道哪个进程是叶子，哪个进程是父进程。为此，我需要在每个过程中执行不同的操作，而我想不出一种方法，
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 |
但我希望输出为：
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
所以树看起来像：
| a |
/ \
| b | | c |
/ / | \
| d | | e | | f | | g |
。
答案 0 :(得分：2)
我们需要做的是跟踪列表中我们处于哪个进程以及列表中子进程位于何处。以下代码显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);
}
示例输出：
Process 0 has pid 2648 and parent 2641. Process 1 has pid 2649 and parent 2648. Process 2 has pid 2650 and parent 2648. Process 3 has pid 2651 and parent 2649. Process 4 has pid 2652 and parent 2650. Process 5 has pid 2653 and parent 2650. Process 6 has pid 2654 and parent 2650.