Question

鉴于以下代码，你能弄明白导致“你输入7个字符”出现了3次，特别是上一次？

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#include <semaphore.h>
void *thread_function(void *arg);
sem_t bin_sem;
#define WORK_SIZE 1024
char work_area[WORK_SIZE];
int main(){
    int res;
    pthread_t a_thread;
    void *thread_result;
    res = sem_init(&bin_sem,0,0);
    if (res!=0){
        perror("Semaphore initialization failed");
        exit(EXIT_FAILURE);
    }
    res = pthread_create(&a_thread,NULL,thread_function,NULL);
    if (res!=0){
        perror("Thread creation failed");
        exit(EXIT_FAILURE);
    }
    printf("Input some text. Enter ‘end’ to finish");
    while (strncmp("end",work_area,3)!=0){
        if(strncmp(work_area,"FAST",4)==0){
            sem_post(&bin_sem);
            strcpy(work_area,"Wheeee...");
        }else{
            fgets(work_area,WORK_SIZE,stdin);
        }
        sem_post(&bin_sem);
    }
    printf("\nWaiting for thread to finish\n");
    res = pthread_join(a_thread,&thread_result);
    if(res!=0){
        perror("Thread join failed!");
        exit(EXIT_FAILURE);
    }
    printf("Thread joined\n");
    sem_destroy(&bin_sem);
    exit(EXIT_SUCCESS);
}
void *thread_function(void* arg){
    sem_wait(&bin_sem);
    while(strncmp("end",work_area,3)!=0){
        printf("You input %d characters\n",strlen(work_area-1));
        sem_wait(&bin_sem);
    }
    pthread_exit(NULL);
}

测试输入/输出：

$ cc -D_REENTRANT thread3a.c -o thread3a -lpthread
$ ./thread3a
Input some text. Enter ‘end’ to finish
Excession
You input 9 characters
FAST
You input 7 characters
You input 7 characters
You input 7 characters
end
Waiting for thread to finish...
Thread joined

Answer 1

输入“FAST”后：

sem_post（while循环结束时的那个） - ＆gt; bin-sem = 1
测试work_area == end（FALSE）
测试work_area == FAST（TRUE）
sem_post - ＆gt; bin-sem = 2
work_area = Wheeee ......
sem_post（while循环结束时的那个） - ＆gt; bin-sem = 3
测试work_area == end（FALSE）
测试work_area = FAST（FALSE）
等待输入

我认为主线程具有优先级，直到它调用fgets（.. stdin ..）时被阻塞，然后线程函数可以运行并使用信号量令牌。

这里有一些发生的事情。

> Input some text.
  Main thread :
    work_area = ?;
    bin_sem = 0;

  thread function :
    wait on semaphore;

< Excession
  Main thread :
    work_area = Excession;
    bin_sem = 1;

  thread function :
    work_area == Excession; (!= end)
> You input 9 characters; (1)
    bin_sem = 0;
    wait on semaphore;

  Main thread :
    work_area == Excession; (!= end)
    work_area == Excession; (!= FAST)
< FAST
    bin_sem = 1;
    work_area == FAST; (!= end)
    work_area == FAST;
    bin_sem = 2;
    work_area == Wheeee...;
    bin_sem = 3;
    wait on stdin;

  thread function :
    work_area == Wheeee...; (!= end)
> You input 7 characters; (Why seven?)
    bin_sem = 3-1 = 2;
> You input 7 characters; (Why seven?)
    bin_sem = 2-1 = 1;
> You input 7 characters; (Why seven?)
    bin_sem = 1-1 = 0;
    wait on semaphore;

Answer 2

这一行在线程函数中是错误的。

printf("You input %d characters\n",strlen(work_area-1));

应该是strlen(work_area)，而不是strlen(work_area-1)。

Answer 3

输出似乎不是来自列出的代码版本：

printf（“输入一些文字”）;

无法生产：

输入一些文字。输入'end'完成

此外，你或多或少地分裂了两个线程之间不正常地控制循环，然后用全局变量绑定它们。这是一种“goto”式结构，只是在寻找麻烦......

保罗。

关于信号量的问题

3 个答案: