叉树C，子进程PID
[英]fork tree C, child processes PID
问题描述
I have to implement fork tree as in this picture . 
我必须实现这张图中的 fork树。
+----+
+--------| P1 |---------+
/ +----+ \
/ / \ \
+----+ +----+ +----+ +----+
| P2 | | P3 | | P4 | | P5 |
+----+ +----+ +----+ +----+
/ \ / \
+----+ +----+ +----+ +----+
| P6 | | P7 | | P8 | | P9 |
+----+ +----+ +----+ +----+
|
+-----+
| P10 |
+-----+
Output must return something like: 输出必须返回如下内容：
P1: pid: 1337 ppid: 1336 child processes pids: 1338, 1339, 1340, 1341
P2: pid: 1338 ppid: 1337 child processes pids: 1342, 1342... etc.
where the child processes pids are pids of every child process of PX, and I don't know how to go about it. 子进程pid是每个PX子进程的pid，我不知道该如何处理。 Can you give me some advie? 你能给我一些建议吗？ My code so far is below, it creates the tree correctly imo, the child processes are a problem. 到目前为止，我的代码在下面，它正确地创建了imo树，子进程是一个问题。
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
int main(int argc, char *argv[])
{
int pidp1;
int pidp2;
switch (pidp1 = fork())
{
case 0:
printf("P2: pid: %d ppid: %d\n", getpid(), getppid());
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
printf("P1: pid: %d ppid: %d child processes pids: %d\n", getpid(), getppid(), pidp1);
wait(NULL);
switch (pidp1 = fork())
{
case 0:
printf("P3: pid: %d ppid: %d\n", getpid(), getppid());
switch (pidp1 = fork())
{
case 0:
printf("P6: pid: %d ppid: %d \n", getpid(), getppid());
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
switch (pidp1 = fork())
{
case 0:
printf("P7: pid: %d ppid: %d\n", getpid(), getppid());
switch (pidp1 = fork())
{
case 0:
printf("P10: pid: %d ppid: %d\n", getpid(), getppid());
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
exit (0);
}
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
exit (0);
}
exit (0);
}
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
switch (pidp1 = fork())
{
case 0:
printf("P4: pid: %d ppid: %d\n", getpid(), getppid());
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
switch (pidp1 = fork())
{
case 0:
printf("P5: pid: %d ppid: %d\n", getpid(), getppid());
switch (pidp1 = fork())
{
case 0:
printf("P8: pid: %d ppid: %d\n", getpid(), getppid());
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
switch (pidp1 = fork())
{
case 0:
printf("P9: pid: %d ppid: %d\n", getpid(), getppid());
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
exit (0);
}
exit (0);
}
exit (0);
break;
case -1:
printf("Blad funkcji\n");
exit (1);
default:
wait(NULL);
exit (0);
}
exit (0);
}
exit (0);
}
exit (0);
}
}
3 个解决方案
解决方案1
2 已采纳 2017-11-08 20:25:11
Seeing as how this is homework, I'll try to give you some pointers. 看到这是家庭作业，我将尝试为您提供一些指导。
Wow, that's a lot of code! 哇，好多代码！ There might be an opportunity here to take some of the code and put it into a function. 这里可能有机会采用一些代码并将其放入函数中。
The tree you've linked is a bit weird-looking (not symmetrical), and it starts counting at 1, which is annoying. 您链接的树看起来有些古怪（不对称），并且从1开始计数，这很烦人。 Nonetheless, we could describe the tree with two arrays: the number of children for each process, and the first number of its leftmost child. 尽管如此，我们仍可以用两个数组来描述树：每个进程的子代数，以及它最左边的子代的第一个数。
/* The number of children for P0, P1, P2, P3, ... */
const int num_children[] = { -1, 4, 0, 2, 0, 2, 0, 1, 0, 0, 0 };
/* The P-number of the first child for P0, P1, P2, P3, ... */
const int first_child[] = { -1, 2, -1, 6, -1, 8, -1, 10, -1, -1, -1 };
You should be able to recreate the tree from your picture using only this information, and without actually looking at the picture. 您应该仅使用此信息就可以从图片中重新创建树，而无需实际查看图片。 (Really, do it - I might have made a mistake.) （真的，这样做-我可能犯了一个错误。）
Then you could write a function that takes a P number, and does the following: 然后，您可以编写一个带有P号的函数，并执行以下操作：
- Look up the number of children we have to make in
num_children.查找我们必须在num_children多少个孩子。 - Look up the P number of the first child in
first_child.在first_child查找第一个孩子的P号。 - Fork a few times, each time running our function in the child, with the right P number, and storing the child's pid in the parent somewhere. 分叉几次，每次以正确的P号在子级中运行我们的函数，并将子级的pid存储在父级中的某个位置。
- Output the line you're describing in your question. 输出您在问题中描述的行。 You've stored all the child pids already, so that should be easy. 您已经存储了所有子pid，因此应该很容易。
- Wait for its own children to complete. 等待自己的孩子完成。
Launch the whole thing with yourfunction(1) . 用yourfunction(1)启动整个过程。
The leaf functions do not have children, so they would wait forever. 叶子功能没有子代，因此它们将永远等待。 That means your program does not quit anymore. 这意味着您的程序不再退出。 Ever. 曾经 This is actually good news: it means you can use Linux tools like ps , pstree , and htop to make sure your tree has the right shape. 这实际上是个好消息：这意味着您可以使用ps ， pstree和htop类的Linux工具来确保树的形状正确。
(The lines of output will probably be output in a random order - I hope that's okay.) （输出行可能会以随机顺序输出-我希望可以。）
Good luck! 祝好运！
解决方案2
1 2017-11-08 20:18:40
Your program is not really readable. 您的程序不是真正可读的。 I think that you should start using structural programming. 我认为您应该开始使用结构化编程。 ie 即
void p2() {
exit(0);
}
void p3() {
if (fork() == 0) p6();
...
exit(0);
}
void p1() {
if (fork() == 0) p2();
if (fork() == 0) p3();
...
exit(0);
}
int main() {
if (fork() == 0) p1();
wait();
...
}
解决方案3
0 2017-11-09 08:53:55
if you wish to having common parent for all child then go for this code instead of writing fork() no of times : 如果您希望所有孩子都有一个共同的父母，那么请选择以下代码，而不要多次编写fork（）：
int main()
{
int i, j, k;
for (i=0; i<10; i++) {
if (fork()==0) {
printf("child %d -> pid : [%d] and ppid : [%d]\n",
i+1, getpid(), getppid());
break;
} else {
if (i==0) {
printf("parent pid : [%d] \n", getpid());
}
wait(0);
}
}
return 0;
}
if you wish to having parent->child1->child2 ie child2 parent is child1 like this sequence then go for this code : 如果您希望让parent-> child1-> child2即child2 parent是child1，例如此序列，请使用以下代码：
int main()
{
int i, j, k;
for (i=0; i<10; i++) {
if (fork()==0) {
printf("child %d -> pid : [%d] and ppid : [%d]\n",
i+1, getpid(), getppid());
} else {
if (i==0) {
printf("parent pid : [%d] \n", getpid());
}
break;
}
}
return 0;
}
Or according to your requirement you can add some condition inside loop. 或者根据您的要求，您可以在循环内添加一些条件。 Hope this will help 希望这会有所帮助
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