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c - 为什么我在 C-Posix 代码中遇到这个信号量问题?

转载 作者:行者123 更新时间:2023-11-30 16:29:02 25 4
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程序必须这样做:

进程 P0 创建 P1 和 P2;

sizeof(buffer) = N(使用命令行插入);

P1 在缓冲区的前 N/2 个元素中插入随机值(N 由用户插入)

P2 在缓冲区的第二部分插入随机值

之后:P1 反转缓冲区的第二部分,然后进程 P0 打印其中的所有元素

如果用户按 CTRL+C ---> 打印缓冲区元素并终止所有进程;

我不知道为什么,但进程 P1 仍然处于暂停状态。我在并发进程 P2 中调用了信号量值(“semaphore_inv”)的增加,并且等待必须将其减少到 0。因此,程序无法正常工作。

#include <stdio.h>
#include <string.h>
#include <signal.h>
#include <unistd.h>
#include <stdlib.h>
#include <semaphore.h>
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <errno.h>

int N;
int buff[1024];

void print(int sig) {
int i;

for(i=0; i<N; i++) {
printf("Slot %d of the buffer is %d\n",i, buff[i]);
}
kill(0,SIGKILL);
}

void main (int argc, char* argv[]) {
int p1, p2;

sem_t semaphore_inv;
sem_t semaphore_read;
sem_t semaphore_write;

struct sembuf sembuf;

N=atoi(argv[1]);

if (N<=0 || N>=1024) {
printf("Inserirt a value > 0 and < 1024\n");
exit(-1);
}

if (argc!=2) {
printf("Insert com N\n");
exit(1);
}

int buffer[N];

//I insert this type of semaphore only to try it
int sem_write = semget(IPC_PRIVATE,1,IPC_CREAT|0666);
if (sem_write <0) printf("Error in the semaphore creation\n");

int sem_write_b = sem_init(&semaphore_write,1,1);
if (sem_write_b<0) printf("Error in the semaphore creation\n");
int sem_inv = sem_init(&semaphore_inv, 1, 0);
if (sem_inv<0) printf("Error in the semaphore creation\n");
int sem_read = sem_init(&semaphore_read,1,0);
if (sem_read<0) printf("Error in the semaphore creation\n");

int ret = semctl(sem_write, 0, SETVAL, 1);
if (ret == -1) printf("Error: semctl, with errno %s\n", strerror(errno));

signal(SIGINT, print);

p1 = fork();
p2 = fork();

if (p1 < 0) {
printf("P1: error, fork\n");
exit(-2);
}

if (p2 < 0) {
printf("P2: error, fork\n");
exit(-2);
}

if (p1==0) {
loop:
sembuf.sem_num=0;
sembuf.sem_op= -1;
sembuf.sem_flg=0;
semop(sem_write, &sembuf, 1);
int i;
for (i=0; i<N/2; i++) {
buffer[i] = rand();
printf("P1: the insert value in buffer[%d] is %d\n",i , buffer[i]);
}

sem_wait(&semaphore_inv);
printf("P1: i'm going to invert the second part of the buffer\n");
int j=1;
for (i=N/2; i<N; i++){
int buffer_prev;
buffer_prev=buffer[i];
buffer[i] = buffer[N-j];
buffer[N-j] = buffer_prev;
j++;
}
sem_post(&semaphore_read);
sleep(1);
goto loop;
}

if (p2==0) {
loop_b:
sem_wait(&semaphore_write);
int i;
for (i=N/2; i<N; i++) {
buffer[i] = rand();
printf("P2: the value insert in buffer[%d] is %d\n", i, buffer[i]);
}
sem_post(&semaphore_inv);
sleep(1);
goto loop_b;
}

else{
sem_wait(&semaphore_read);
int k;
for (k=0; k<N; k++) {
buff[k] = buffer[k];
printf(" slot %d of the buffer is %d\n", buffer[k]);
}

sem_post(&semaphore_write);

sembuf.sem_num=0;
sembuf.sem_op= +1;
sembuf.sem_flg=0;
semop(sem_write, &sembuf, 1);
}

}

最佳答案

涉及四个流程。插图:

<小时/>
#include <stdio.h>
#include <unistd.h>

int main(void)
{
int pid = -2, pid1 = -2, pid2 = -2;


pid1 = fork();
pid2 = fork();
mypid = getpid();

printf("Pid= {%d, %d %d}\n", mypid, pid1,pid2);

return 0;
}

关于c - 为什么我在 C-Posix 代码中遇到这个信号量问题?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52095312/

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