sem_wait在我的程序中被忽略

Question

我在linux中編寫了一個簡單的項目用於共享內存。兩個程序共享內存，一個是給它寫信，另一個是從它讀取它們。我決定使用信號量來確保在讀取之前不會產生新的字母。

問題是我的編寫器進程忽略了sem_wait（讀取），當它的值是0並且它應該等待。它甚至在讀者開始之前完成它的工作。我運行它通過./writer & ./reader。

我附上了代碼。這裏有一些未使用的元素，因爲它還不是最終版本。然而，問題已經出現。

/* writer.c */ 
#include <stdio.h> 
#include <stdlib.h> 
#include <sys/shm.h> 
#include <unistd.h> 
#include <stdlib.h> 
#include <semaphore.h> 

int main(int argc, char *argv[]) 
{ 
    key_t shmkey = 0xF00; 
    int bytes = sizeof(char)*3 + sizeof(sem_t) * 3; 
    int shmid; 
    char* sharedMemory; 
    sem_t *writing, *reading, *working; 

    if ((shmid = shmget(shmkey, bytes, IPC_CREAT | IPC_EXCL | 0666)) < 0) 
    { 
     shmdt((void*) sharedMemory); 
     shmctl(shmid, IPC_RMID, NULL); 
     return 1; 
    } 
    if ((sharedMemory = (char*) shmat(shmid, NULL, 0)) == (char*) -1) 
    { 
     shmdt((void*) sharedMemory); 
     shmctl(shmid, IPC_RMID, NULL); 
     return 1; 
    } 

    writing = (sem_t*)(sharedMemory + 3); 
    reading = writing + 1; 
    working = reading + 1; 

    sem_init(writing, 0, 0); 
    sem_init(reading, 0, 0); 

    sharedMemory[2] = 'w'; // writer is running 
    char c; 
    for(c = 'a'; c <= 'z'; ++c) 
    { 
     *sharedMemory = c; 
     sem_post(writing); 
     sem_wait(reading); 
    } 
    sharedMemory[2] = 'q'; 
    while (sharedMemory[2] != 'w'); 
    sharedMemory[2] = 'q'; 
    shmdt((void*) sharedMemory); 
    shmctl(shmid, IPC_RMID, NULL); 
    return 0; 
} 

與讀者，

/* reader.c */ 
#include <stdio.h> 
#include <stdlib.h> 
#include <sys/shm.h> 
#include <unistd.h> 
#include <stdlib.h> 
#include <semaphore.h> 

int main(int argc, char *argv[]) 
{ 
    key_t shmkey = 0xF00; 
    int bytes = sizeof(char)*3 + sizeof(sem_t) * 3; 
    int shmid; 
    char* sharedMemory; 
    sem_t *writing, *reading, *working; 

    sleep(1); // wait until writer allocates fresh memory 
    if ((shmid = shmget(shmkey, bytes, 0666)) < 0) 
    { 
     shmdt((void*) sharedMemory); 
     return 1; 
    } 
    if ((sharedMemory = (char*) shmat(shmid, NULL, 0)) == (char*) -1) 
    { 
     shmdt((void*) sharedMemory); 
     return 1; 
    } 

    if (sharedMemory[2] != 'w') // is writer running? 
    { 
     shmdt((void*) sharedMemory); 
     return 1; 
    } 

    writing = (sem_t*)(sharedMemory + 3); 
    reading = writing + 1; 
    working = reading + 1; 

    //sleep(5); //@REMOVE 

    char c; 
    do 
    { 
     sem_wait(writing); 
     c = *sharedMemory; 
     sem_post(reading); 
     printf("%c\n", c); 
    } while (sharedMemory[2] == 'w'); 
    sharedMemory[2] = 'w'; 
    shmdt((void*) sharedMemory); 
    return 0; 
} 

Answer 1

sharedMemory + 3沒有正確的sem_t類型一致。由於您不知道sem_t的對齊要求，因此您需要確保sem_t對象在共享內存段的偏移量爲sizeof(sem_t)的倍數處開始（這是可行的，因爲任何對象的對齊要求均勻分割其大小）。

請注意，您應該檢查sem_wait和sem_post的返回值。然後你可以檢查errno，如果他們失敗了，那會給你提供他們爲什麼失敗的信息（但是在你的情況下，我懷疑errno的值可能不如幫助）。