我有一个分配,该赋值使我将此代码转换为使父进程等待所有子进程完成的代码。 PS:第一个代码有4个过程,需要使用waitpid来解决。
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>
int main(){
pid_t p = fork();
pid_t k = fork();
if(p>0){
printf("p=%d: PID = %d\n", p, getpid());
sleep(45);
exit(0);
}
else if(p==0){
printf("p=%d: PID = %d\n", p, getpid());
exit(0);
}
else if(p<0){
printf("ERRO! p=%d\n", p);
exit(p);
}
}
我已经尝试过了,但是我认为这仅适用于1个子进程,不适用于许多子进程。
int main(){
pid_t p = fork();
pid_t k = fork();
if(p<0){
printf("fodeu");
exit(p);
}
else if(p==0){
printf("");
exit(0);
}
else{
for(i=0;i<4;i++){
int returnstatus;
waitpid(p,&returnstatus,0);
if(returnstatus == 0){
printf("o processo filho correu normalmente");
}
else if(returnstatus == 1){
printf("o processo filho ardeu");
}
}
}
}
答案 0 :(得分:0)
这是一种实现方法;还有很多其他人。
#include <stdio.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <unistd.h>
static void wait_for_kids(void);
int main(void)
{
pid_t p = fork();
pid_t k = fork();
if (p > 0)
{
printf("p=%d: PID = %d\n", p, getpid());
sleep(5);
wait_for_kids();
printf("%d: p = %5d, k = %5d - exiting\n", getpid(), p, k);
exit(0);
}
else if (p == 0)
{
printf("p=%d: PID = %d\n", p, getpid());
wait_for_kids();
printf("%d: p = %5d, k = %5d - exiting\n", getpid(), p, k);
exit(0);
}
else
{
printf("ERRO! p=%d\n", p);
wait_for_kids();
printf("%d: p = %5d, k = %5d - exiting\n", getpid(), p, k);
exit(p);
}
/*NOTREACHED*/
}
static void wait_for_kids(void)
{
int corpse;
int status;
int pid = getpid();
while ((corpse = waitpid(0, &status, 0)) > 0)
printf("%d: child %d exited with status 0x%.4X\n", pid, corpse, status);
}
示例输出:
p=43445: PID = 43444
p=43445: PID = 43446
p=0: PID = 43445
p=0: PID = 43447
43447: p = 0, k = 0 - exiting
43445: child 43447 exited with status 0x0000
43445: p = 0, k = 43447 - exiting
43446: p = 43445, k = 0 - exiting
43444: child 43445 exited with status 0x0000
43444: child 43446 exited with status 0x0000
43444: p = 43445, k = 43446 - exiting
答案 1 :(得分:0)
这不会完成您的任务,但我希望这对您有所帮助
你去。作业似乎是fork()
的一个谜,您的
老师很有品味:-)
fork()
是不同的。它返回两次。
getpid()
实际上,任务不是很好。通常使用`fork() 永远不要让任何分支逃逸到封闭的代码中以避免完成 废话。像这样
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
int main()
{
pid_t pid = fork();
if (pid == 0 /*child*/) {
printf("PID %d (child) doing work\n", pid);
sleep(5);
exit(0); // don't let it continue (leak) into parent code
}
else if (pid > 0 /*parent*/) {
int status;
pid_t terminated;
printf("PID %d (parent) waiting for child PID %d\n", getpid(), pid);
terminated = waitpid(pid, &status, 0);
if (terminated == -1) {
perror("waitpid");
exit(1);
}
if (WIFEXITED(status))
printf("child exited normally with status %d\n", WEXITSTATUS(status));
else
printf("hm. child died otherwise. see 'man waidpid' for more\n");
}
return 0;
}
牢记这一点,看看这两条无辜的表情,
pid_t p = fork(); // two processes after this
pid_t k = fork(); // executed by **two** processes, again duplicating
因此,在这两行之后,我们有四个进程执行其余的
并行代码。这是大脑爆炸的地方。什么
问k
的值时,p
行的泄漏子代会做什么
是吗?
看看这个小程序的输出,看看它的作用是什么 泄漏。
#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
int main(){
printf("MAIN PID %d\n", getpid());
fork();
fork();
printf("PID %d, PPID %d\n", getpid(), getppid());
return 0;
}