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在C语言中使用条件变量和互斥体来同步线程

wait mutex signals pthreads c
  •  0
  • Don  · 技术社区  · 7 年前

    文件一 

    h
0
h
0
h
0
h
0
h
0

    文件二 

    0
i
0
i
0
i
0
i
0
i

    我正在尝试使用pthread mutex lock/unlock以及signal,并等待这项工作。然而,我一直处于僵局状态。

    二 螺纹。目前,它们相互镜像意味着它们做相同的事情,只是使用不同的文件和相反的条件。

    线程示例: 

    char final[1001];
pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t condition1 = PTHREAD_COND_INITIALIZER;
pthread_cond_t condition2 = PTHREAD_COND_INITIALIZER;

int w = 1;

void *get()
{
//start reading
while (count < //number)
{
    pthread_mutex_lock(&lock);

    //read line

    //if we've reached a zero
    {
        w = 2;             

         while(w == 2)
         {
            pthread_cond_wait(&condition1, &lock);
         }


         pthread_mutex_unlock(&lock);
    }
    else
    {   
       if(w == 1)
       {

            if(strlen(placeHolderChars)>0)
            {
                 placeHolderChars[1] = '\0';
            }

            //copy char to array

            w= 2;
            pthread_cond_signal(&condition2);
            pthread_mutex_unlock(&lock);
        }
    }

    if(feof(file))
    {     
        fclose(file);
        break;
    }
    count++;

 }

return 0;
}

    更新:在使用较大的文件时,先信号后等待策略实际上不起作用。还在努力!

    2 回复  |  直到 7 年前
        1
  •  0
  •   Jonathan Leffler    7 年前

    此代码似乎有效。主要的重大变化是增加 pthread_cond_signal() 在进入 while (who == N) 循环。

    其他更改包括基本的调试打印,以便于查看哪个线程正在做什么。请注意,调试消息以换行结束。 

    #include <assert.h>
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

extern void *getMessage1(void *arg);
extern void *getMessage2(void *arg);

static char message[4096];

int main(void)
{
    pthread_t id1;
    pthread_t id2;

    pthread_create((&id1), NULL, getMessage1, NULL);
    pthread_create((&id2), NULL, getMessage2, NULL);

    pthread_join(id1, NULL);
    pthread_join(id2, NULL);

    for (int j = 0; j < 1001 && message[j] != '\0'; j++)
        printf("%c ", message[j]);
    putchar('\n');

    return 0;
}

static pthread_mutex_t lock = PTHREAD_MUTEX_INITIALIZER;
static pthread_cond_t condition1 = PTHREAD_COND_INITIALIZER;
static pthread_cond_t condition2 = PTHREAD_COND_INITIALIZER;

static int who = 1;

void *getMessage1(void *arg)
{
    assert(arg == NULL);
    const char filename[] = "Student1";
    FILE *studentOne = fopen(filename, "r");
    if (studentOne == NULL)
    {
        fprintf(stderr, "Failed to open file %s for reading\n", filename);
        exit(EXIT_FAILURE);
    }

    size_t howManyChars;
    char *placeHolderChars;
    int count = 1;

    while (count < 501)
    {
        placeHolderChars = NULL;
        if (getline(&placeHolderChars, &howManyChars, studentOne) == -1)
            break;
        printf("M1(%d): [%s]\n", count, placeHolderChars);
        pthread_mutex_lock(&lock);
        if (strcmp(placeHolderChars, "0\n") == 0)
        {
            printf("M1: Two's turn - 1\n");
            pthread_cond_signal(&condition2);
            who = 2;
            while (who == 2)
            {
                pthread_cond_wait(&condition1, &lock);
            }
            free(placeHolderChars);
        }
        else
        {
            if (who == 1)
            {
                if (strlen(placeHolderChars) > 0)
                {
                    placeHolderChars[1] = '\0';
                }
                strcat(message, placeHolderChars);
                free(placeHolderChars);
                who = 2;
                pthread_cond_signal(&condition2);
            }
            else
                printf("M1: Two's turn - 2\n");
        }
        pthread_mutex_unlock(&lock);
        count++;
    }

    fclose(studentOne);
    return 0;
}

void *getMessage2(void *arg)
{
    assert(arg == NULL);
    const char filename[] = "Student2";
    FILE *studentTwo = fopen(filename, "r");
    if (studentTwo == NULL)
    {
        fprintf(stderr, "Failed to open file %s for reading\n", filename);
        exit(EXIT_FAILURE);
    }

    size_t howManyChars;
    char *placeHolderChars;
    int count = 0;

    while (count < 501)
    {
        placeHolderChars = NULL;
        if (getline(&placeHolderChars, &howManyChars, studentTwo) == -1)
            break;
        printf("M2(%d): [%s]\n", count, placeHolderChars);
        pthread_mutex_lock(&lock);
        if (strcmp(placeHolderChars, "0\n") == 0)
        {
            printf("M2: One's turn - 1\n");
            pthread_cond_signal(&condition1);
            who = 1;
            while (who == 1)
            {
                pthread_cond_wait(&condition2, &lock);
            }
            free(placeHolderChars);
        }
        else
        {
            if (who == 2)
            {
                if (strlen(placeHolderChars) > 0)
                {
                    placeHolderChars[1] = '\0';
                }
                strcat(message, placeHolderChars);
                free(placeHolderChars);
                who = 1;
                pthread_cond_signal(&condition1);
            }
            else
                printf("M2: One's turn - 2\n");
        }
        pthread_mutex_unlock(&lock);
        count++;
    }

    fclose(studentTwo);
    return 0;
}

    您应该能够优化代码,以便将包含相关每线程数据(文件名、当前线程条件、其他线程条件,可能是“线程ID”)的结构传递给单个函数,因此您只需要 getMessage()

    输出: 

    M1(1): [h
]
M2(0): [0
]
M1(2): [0
]
M2: One's turn - 1
M1: Two's turn - 1
M2(1): [i
]
M2(2): [0
]
M2: One's turn - 1
M1(3): [h
]
M1(4): [0
]
M1: Two's turn - 1
M2(3): [i
]
M2(4): [0
]
M2: One's turn - 1
M1(5): [h
]
M1(6): [0
]
M1: Two's turn - 1
M2(5): [i
]
M2(6): [0
]
M2: One's turn - 1
M1(7): [h
]
M1(8): [0
]
M1: Two's turn - 1
M2(7): [i
]
M2(8): [0
]
M2: One's turn - 1
M1(9): [h
]
M1(10): [0
]
M1: Two's turn - 1
M2(9): [i
]
h i h i h i h i h i

    我对这段代码不太满意。正如我所暗示的那样,我创建了一个修改版本,其中包含两个线程使用的单个函数,并修改了读取的行的打印,以避免打印换行符(使输出更加紧凑)。有时并非所有时候,它都会在最后陷入僵局。两个样本记录道,一个工作,一个死锁(程序名 pth47 

    $ pth47
M2(1): [0]
M2: 1's turn - 1
M1(1): [h]
M1(2): [0]
M1: 2's turn - 1
M2(2): [i]
M2(3): [0]
M2: 1's turn - 1
M1(3): [h]
M1(4): [0]
M1: 2's turn - 1
M2(4): [i]
M2(5): [0]
M2: 1's turn - 1
M1(5): [h]
M1(6): [0]
M1: 2's turn - 1
M2(6): [i]
M2(7): [0]
M2: 1's turn - 1
M1(7): [h]
M1(8): [0]
M1: 2's turn - 1
M2(8): [i]
M2(9): [0]
M2: 1's turn - 1
M1(9): [h]
M1(10): [0]
M1: 2's turn - 1
M2(10): [i]
h i h i h i h i h i 
$ pth47
M1(1): [h]
M2(1): [0]
M1(2): [0]
M2: 1's turn - 1
M1: 2's turn - 1
M2(2): [i]
M2(3): [0]
M2: 1's turn - 1
M1(3): [h]
M1(4): [0]
M1: 2's turn - 1
M2(4): [i]
M2(5): [0]
M2: 1's turn - 1
M1(5): [h]
M1(6): [0]
M1: 2's turn - 1
M2(6): [i]
M1(7): [h]
M1(8): [0]
M2(7): [0]
M1: 2's turn - 1
M2: 1's turn - 1
M1(9): [h]
M1(10): [0]
M2(8): [i]
M1: 2's turn - 1
M2(9): [0]
M2: 1's turn - 1
^C
$

    我还没有找到异常现象。这并不像“线程一个先走”那么简单;有一些例子是线程一先完成的,它完成得很好。

        2
  •  0
  •   Erik Alapää    7 年前

    我认为提出这个问题的人做了两次,有点恼人。FWIW,这是我对副本的回答: Pthread Synchronization: Back & Forth Reading of Two Text Files

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