[英]How many processes will be created
我有这段代码,想知道将创建多少个进程。 我不确定,因为我认为循环将是 12 个进程,但也可能是 8 个。
#include <unistd.h>
#include <sys/types.h>
int main() {
pid_t childpid;
int i;
childpid = fork();
for (i = 0; i < 3 && childpid == 0; i++) {
if (childpid == -1) {
perror("Failed to fork.");
return 1;
}
fprintf(stderr, "A\n");
childpid = fork();
if (childpid == 0) {
fprintf(stderr, "B\n");
childpid = fork();
fprintf(stderr, "C\n");
}
}
return 0;
}
是的,您可以使用 aptitude 来评估创建了多少进程,但我让程序为我决定。
子进程和父进程使用semaphore
同步,并且使用pcount
变量来跟踪已创建进程的数量。 每当pcount
被评估为零时, childpid
就会增加。 以下是对您的程序的补充。
#include <semaphore.h>
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/mman.h>
int main(void)
{
/* Initialize and setup a semaphore */
sem_t* sema = mmap(NULL, sizeof(sem_t), PROT_READ | PROT_WRITE, MAP_SHARED | MAP_ANONYMOUS, -1, 0);
if (sema == MAP_FAILED)
exit(1);
if (sem_init(sema, 1, 1) != 0)
exit(1);
/* Initialize pcount */
int* pcount = mmap(NULL, sizeof(int), PROT_READ | PROT_WRITE, MAP_SHARED | MAP_ANONYMOUS, -1, 0);
if (pcount == MAP_FAILED)
exit(1);
*pcount = 1;
printf("pcount = %d\n", *pcount);
pid_t childpid;
int i;
childpid = fork();
if (childpid == 0) {
sem_wait(sema);
*pcount = *pcount + 1;
printf("pcount = %d\n", *pcount);
sem_post(sema);
}
for (i = 0; i < 3 && childpid == 0; i++) {
if (childpid == -1) {
perror("Failed to fork.");
return 1;
}
fprintf(stderr, "A\n");
childpid = fork();
if (childpid == 0) {
sem_wait(sema);
*pcount = *pcount + 1;
printf("pcount = %d\n", *pcount);
sem_post(sema);
fprintf(stderr, "B\n");
childpid = fork();
if (childpid == 0) {
sem_wait(sema);
*pcount = *pcount + 1;
printf("pcount = %d\n", *pcount);
sem_post(sema);
}
fprintf(stderr, "C\n");
}
}
return 0;
}
端子 Session:
$ gcc SO.c -lpthread
$ ./a.out
pcount = 1
pcount = 2
A
pcount = 3
B
C
pcount = 4
C
A
pcount = 5
B
C
pcount = 6
C
A
pcount = 7
B
C
pcount = 8
C
所以是的, 8
是答案。 十分简单:)
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