我有这段代码应该创建三个子进程,每个子进程将执行一个小的数学运算。然后, parent 应该使用所有子进程的结果并得到最终答案,但我找不到一种方法来实际读取 parent 的 child 的结果。有办法做到这一点吗?
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
int main(void)
{
int pid1, pid2, pid3, status;
int a=1, b=2, c=5, d=4, e=6, f=3, g;
int t1, t2, t3;
printf("Hello World!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n");
printf("Here I am before use of forking\n");
printf("I am the PARENT process and pid is : %d\n",getpid());
pid1 = fork( );
if (pid1 == 0)
{
printf("\n\nHere I am just after child forking1\n");
printf("I am the Child process and pid1 is :%d\n",getpid());
printf("My parent's pid is :%d\n",getppid());
t1 = a+b;
printf("The answer for t1 is: %d\n", t1);
exit(0);
}
else
{
wait(&status);
printf("\nHere I am just after parent forking1\n");
printf("I am the Parent process and pid is: %d\n",getpid());
}
pid2 = fork( );
if (pid2 == 0)
{
printf("\n\nHere I am just after child forking2\n");
printf("I am the Child process and pid2 is :%d\n",getpid());
printf("My parent's pid is :%d\n",getppid());
t2 = c+d;
printf("The answer for t2 is: %d\n", t2);
exit(0);
}
else
{
wait(&status);
printf("\nHere I am just after parent forking2\n");
printf("I am the Parent process and pid is: %d\n",getpid());
}
pid3 = fork( );
if (pid3 == 0)
{
printf("\n\nHere I am just after child forking3\n");
printf("I am the Child process and pid3 is :%d\n",getpid());
printf("My parent's pid is :%d\n",getppid());
t3 = e/f;
printf("The answer for t3 is: %d\n", t3);
exit(0);
}
else
{
wait(&status);
printf("\nHere I am just after parent forkingALL\n");
printf("I am the Parent process and pid is: %d\n",getpid());
}
printf("\n\nThe final answer for t1 is: %d\n", t1);
printf("The final answer for t2 is: %d\n", t2);
printf("The final answer for t3 is: %d\n", t3);
g = t1*t2-t3;
printf("The final answer for g is: %d\n", g);
}
最佳答案
您可以使用一种非常简单的技术来做到这一点,即共享内存。我将给出一个完整的例子来说明它是如何工作的。
首先,假设我想编写一个程序打印斐波那契数列中的前n
项(我知道这样做不合逻辑,但是这是一个简单的例子,所以每个人都可以理解)。
- 我有一个父级读取代表前
n
个项的整数值 - 然后父进程将创建一个子进程并将
n
传递给它 - 然后 child 应该计算前 n 项并将它们返回给 parent 。
#include <stdio.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <fcntl.h>
#include <sys/stat.h>
void printFibo(int n, int *fibo)
{
int i;
for(i=0; i<=n; i++)
printf("%d -> %d\n" ,i, fibo[i]);
}
void computeFibo(int n, int *fibo)
{
int i;
fibo[0] = 0;
fibo[1] = 1;
for (i=2; i<=n; i++)
fibo[i] = fibo[i-1] + fibo[i-2];
}
int main(int argc, char *argv[])
{
pid_t childPID;
int status;
int shm_fd;
int* shared_memory;
int msize; // the size (in bytes) of the shared memory segment
const char *name = "FIBONACCI_SERIES";
int n;
if (argc!=2)
{
fprintf(stderr, "usage: %s <Fibonacci number to be generated>\n", argv[0]);
return -1;
}
n = atoi(argv[1]);
if (n < 0)
{
fprintf(stderr, "Illegal fibonacci number: %s\n", argv[1]);
return -2;
}
// calculating the array size based on the number of terms being passed from child to parent
msize = (n+2)*sizeof(int);
// open the memory
shm_fd = shm_open (name, O_CREAT | O_EXCL | O_RDWR, S_IRWXU | S_IRWXG);
if (shm_fd < 0)
{
fprintf(stderr,"Error in shm_open()");
return -3;
}
printf("Created shared memory object %s\n", name);
// attach the shared memory segment
ftruncate(shm_fd, msize);
printf("shmat returned\n");
// allocating the shared memory
shared_memory = (int *) mmap(NULL, msize, PROT_READ | PROT_WRITE, MAP_SHARED, shm_fd, 0);
if (shared_memory == NULL)
{
fprintf(stderr,"Error in mmap()");
return -3;
}
printf("Shared memory segment allocated correctly (%d bytes).\n", msize);
shared_memory[0] = n;
childPID=fork();
if ( childPID == -1 )
{
fprintf(stderr, "Cannot proceed. fork() error");
return -4;
}
if (childPID == 0)
{
// then we're the child process
computeFibo(shared_memory[0],shared_memory+1);
exit(0);
}
else
{
// parent will wait until the child finished
wait(&status);
// print the final results in the
printFibo(shared_memory[0], shared_memory+1);
// now detach the shared memory segment
shm_unlink(name);
}
return 0;
}
关于c - 将值从子进程传递到父进程,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12499745/