C中的Posix信号灯等待和信号循环问题

进一步研究我的代码。在用户的帮助下，我设法在代码中实现了共享内存。但是我遇到了一个问题。此代码仅用于演示缓冲区输入和输出算法。

In wait_t() the while loop keeps going even when sem_wait() is called. The loop should stop when the wait is called when length is under 1. I've tried using break; to stop the loop and when is called sem_post() the length was still 3 instead of 0.

So how can I get the while loop to stop when sem_wait() is called and how to share length across both functions since it uses processes.

#include <stdio.h>
#include <semaphore.h> 
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <sys/mman.h>
#include <sys/ipc.h>
#include <fcntl.h> 
#include <sys/shm.h> 
#include <sys/stat.h> 
sem_t *mutex; 
int length;
int out;
int in;
void wait_t(int a)
{
    while(1){
        if (a < 1)
        {
            sem_wait(&mutex);
            printf("waiting\n");
            //doesn't stop unless break;
        }
        //write to buffer
        a--;
        printf("a\n");
    }
}
void signal_t(int a)
{
    printf("length = %d \n", a);
    while(a != 3){
        if (a < 1)
        {
            sem_post(&mutex);
            printf("signal\n");
        }
        //print to buffer
        printf("b\n");
        a++;
    }
}

int main()
{ 
    length = 3;
    int shared_mem_id = shmget(IPC_PRIVATE, sizeof(sem_t), IPC_CREAT);
    mutex = shmat(shared_mem_id, NULL, 0);
    sem_init(&mutex, 1, 1);
    int i = -2;
    if(fork() == 0){ // create 1st child process
        mutex = shmat(shared_mem_id, NULL, 0);
        wait_t(length);
        exit(0); 
    }
    shmdt(mutex);
    wait(NULL);
    if(fork() == 0){ // create 2nd child process
        mutex = shmat(shared_mem_id, NULL, 0);
        signal_t(length);
        exit(0); 
    }
    shmdt(mutex);
    wait(NULL);
    if(fork() == 0){ // create 3nd child process
        mutex = shmat(shared_mem_id, NULL, 0);
        signal_t(i);
        exit(0); 
    }
    shmdt(mutex);
    wait(NULL);
    printf("%d\n", i);
    exit(0); 
}