Using an all semaphore program instead of mutex and cond

Using an all semaphore program instead of mutex and cond_t

I am a beginner at C and I was just studying more in a book I picked up. I got to mutex locks and conditional variables and stuff like that. I saw an exercise in the book that said to switch a program that had mutex locks and conditional to a program that used semaphores instead to achieve the same results. I'm currently stuck because when I run the code, I believe I have the situation of being "deadlocked" and nothing happens after. I'm not sure what's wrong but any help would be appreciated. This is what I have now.

int main(){

  pthread_t pid;

  pthread_t cid;



  sem_init(&empty, 1, BUFSIZE);

  sem_init(&full, 1, BUFSIZE);



  randData =  (unsigned int) time(NULL);



  printf("Creating threadsnn");



  pthread_create(&pid, NULL, produce, ids);

  pthread_create(&cid, NULL, consume, ids);  

  pthread_join(pid, NULL);

  pthread_join(cid, NULL);

}

edited Nov 27 '18 at 10:09

asked Nov 25 '18 at 8:16

James Williams

Just to clarify -- should there be multiple producers and consumers in the semaphore version? It looks like you had several of each in your condition variable output, but I only see you creating one of each in the new code.

– Paul
Nov 25 '18 at 8:33

There are 2 in produce. 1 for pthread_cond_wait and pthread_cond_signal There are 2 in consume, same as produce sorry about that. where should I be creating more?

– James Williams
Nov 25 '18 at 8:37

add a comment |

int main(){

  pthread_t pid;

  pthread_t cid;



  sem_init(&empty, 1, BUFSIZE);

  sem_init(&full, 1, BUFSIZE);



  randData =  (unsigned int) time(NULL);



  printf("Creating threadsnn");



  pthread_create(&pid, NULL, produce, ids);

  pthread_create(&cid, NULL, consume, ids);  

  pthread_join(pid, NULL);

  pthread_join(cid, NULL);

}

edited Nov 27 '18 at 10:09

asked Nov 25 '18 at 8:16

James Williams

Just to clarify -- should there be multiple producers and consumers in the semaphore version? It looks like you had several of each in your condition variable output, but I only see you creating one of each in the new code.

– Paul
Nov 25 '18 at 8:33

There are 2 in produce. 1 for pthread_cond_wait and pthread_cond_signal There are 2 in consume, same as produce sorry about that. where should I be creating more?

– James Williams
Nov 25 '18 at 8:37

add a comment |

int main(){

  pthread_t pid;

  pthread_t cid;



  sem_init(&empty, 1, BUFSIZE);

  sem_init(&full, 1, BUFSIZE);



  randData =  (unsigned int) time(NULL);



  printf("Creating threadsnn");



  pthread_create(&pid, NULL, produce, ids);

  pthread_create(&cid, NULL, consume, ids);  

  pthread_join(pid, NULL);

  pthread_join(cid, NULL);

}

edited Nov 27 '18 at 10:09

asked Nov 25 '18 at 8:16

James Williams

int main(){

  pthread_t pid;

  pthread_t cid;



  sem_init(&empty, 1, BUFSIZE);

  sem_init(&full, 1, BUFSIZE);



  randData =  (unsigned int) time(NULL);



  printf("Creating threadsnn");



  pthread_create(&pid, NULL, produce, ids);

  pthread_create(&cid, NULL, consume, ids);  

  pthread_join(pid, NULL);

  pthread_join(cid, NULL);

}

c pthreads conditional mutex semaphore

edited Nov 27 '18 at 10:09

asked Nov 25 '18 at 8:16

James Williams

edited Nov 27 '18 at 10:09

asked Nov 25 '18 at 8:16

James Williams

edited Nov 27 '18 at 10:09

asked Nov 25 '18 at 8:16

James Williams

asked Nov 25 '18 at 8:16

James Williams

asked Nov 25 '18 at 8:16

James Williams

Just to clarify -- should there be multiple producers and consumers in the semaphore version? It looks like you had several of each in your condition variable output, but I only see you creating one of each in the new code.

– Paul
Nov 25 '18 at 8:33

There are 2 in produce. 1 for pthread_cond_wait and pthread_cond_signal There are 2 in consume, same as produce sorry about that. where should I be creating more?

– James Williams
Nov 25 '18 at 8:37

add a comment |

Just to clarify -- should there be multiple producers and consumers in the semaphore version? It looks like you had several of each in your condition variable output, but I only see you creating one of each in the new code.

– Paul
Nov 25 '18 at 8:33

There are 2 in produce. 1 for pthread_cond_wait and pthread_cond_signal There are 2 in consume, same as produce sorry about that. where should I be creating more?

– James Williams
Nov 25 '18 at 8:37

Just to clarify -- should there be multiple producers and consumers in the semaphore version? It looks like you had several of each in your condition variable output, but I only see you creating one of each in the new code.

– Paul
Nov 25 '18 at 8:33

There are 2 in produce. 1 for pthread_cond_wait and pthread_cond_signal There are 2 in consume, same as produce sorry about that. where should I be creating more?

– James Williams
Nov 25 '18 at 8:37

add a comment |

2 Answers
2

active

oldest

votes

In code for producer, below, mutex_wait() is for same "mutex" twice which can lock while post() is only once in consumer.

sem_wait(&empty);

sem_wait(&mutex);  <---

while ( isFull() ){

  printf("%*sProducer %d waitsn", id*5, "", id);

}



enter(value);

printf("%*sProducer %d stores %d    ", id*5, "", id, value);

print();

printf("n");



sem_wait(&mutex);  <---

sem_post(&full);

Few ideas to get it good in shape is When there is wait for empty or full, just add one check rather than loop and do mutex wait() in producer and from consumer post() it when it is empty or not full.

answered Nov 25 '18 at 8:38

anand

1377

Can you give me an example of what you mean? Thank you.

– James Williams
Nov 25 '18 at 9:50

add a comment |

It looks like your solution is pretty close, but you still have some of the condition/wait stuff leftover that you shouldn't need. For example, once a thread take the appropriate semaphore, it knows it can proceed.

Check me on this code -- there may be some issues, but here's what I'd recommend (it's pseudo C)

void* produce(void* arg)

{ 

  // Wait for any item to be present in queue

  sem_wait(&isNotFullSem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  enter(value);



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not empty

  //  (probably not the best name, but similar to condition)

  sem_post(&isNotEmptySem);

}



void* consume(void* arg)

{

  // Wait for empty space to be present in queue

  sem_wait(&isNotEmptySem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  value = leave();



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not full

  sem_post(&isNotFullSem);

}

answered Nov 25 '18 at 8:40

Paul

3606

I appreciate the feedback but I keep on getting a deadlock for some reason

– James Williams
Nov 25 '18 at 9:50

@JamesWilliams Hmm, I can't see how the code I have above would result in a deadlock. Can you post your updates to your question so I can take a look? Does the deadlock occur only when the program should terminate? You need some handling for that case (i.e. set exit = 1 and post isNotFull and isNotEmpty to make threads return).

– Paul
Nov 25 '18 at 9:56

put my update using your psuedocode as an answer, thanks

– James Williams
Nov 25 '18 at 13:11

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

In code for producer, below, mutex_wait() is for same "mutex" twice which can lock while post() is only once in consumer.

sem_wait(&empty);

sem_wait(&mutex);  <---

while ( isFull() ){

  printf("%*sProducer %d waitsn", id*5, "", id);

}



enter(value);

printf("%*sProducer %d stores %d    ", id*5, "", id, value);

print();

printf("n");



sem_wait(&mutex);  <---

sem_post(&full);

answered Nov 25 '18 at 8:38

anand

1377

Can you give me an example of what you mean? Thank you.

– James Williams
Nov 25 '18 at 9:50

add a comment |

In code for producer, below, mutex_wait() is for same "mutex" twice which can lock while post() is only once in consumer.

sem_wait(&empty);

sem_wait(&mutex);  <---

while ( isFull() ){

  printf("%*sProducer %d waitsn", id*5, "", id);

}



enter(value);

printf("%*sProducer %d stores %d    ", id*5, "", id, value);

print();

printf("n");



sem_wait(&mutex);  <---

sem_post(&full);

answered Nov 25 '18 at 8:38

anand

1377

Can you give me an example of what you mean? Thank you.

– James Williams
Nov 25 '18 at 9:50

add a comment |

In code for producer, below, mutex_wait() is for same "mutex" twice which can lock while post() is only once in consumer.

sem_wait(&empty);

sem_wait(&mutex);  <---

while ( isFull() ){

  printf("%*sProducer %d waitsn", id*5, "", id);

}



enter(value);

printf("%*sProducer %d stores %d    ", id*5, "", id, value);

print();

printf("n");



sem_wait(&mutex);  <---

sem_post(&full);

answered Nov 25 '18 at 8:38

anand

1377

In code for producer, below, mutex_wait() is for same "mutex" twice which can lock while post() is only once in consumer.

sem_wait(&empty);

sem_wait(&mutex);  <---

while ( isFull() ){

  printf("%*sProducer %d waitsn", id*5, "", id);

}



enter(value);

printf("%*sProducer %d stores %d    ", id*5, "", id, value);

print();

printf("n");



sem_wait(&mutex);  <---

sem_post(&full);

answered Nov 25 '18 at 8:38

anand

1377

answered Nov 25 '18 at 8:38

anand

1377

answered Nov 25 '18 at 8:38

anand

1377

answered Nov 25 '18 at 8:38

anand

1377

Can you give me an example of what you mean? Thank you.

– James Williams
Nov 25 '18 at 9:50

add a comment |

Can you give me an example of what you mean? Thank you.

– James Williams
Nov 25 '18 at 9:50

Can you give me an example of what you mean? Thank you.

– James Williams
Nov 25 '18 at 9:50

add a comment |

Check me on this code -- there may be some issues, but here's what I'd recommend (it's pseudo C)

void* produce(void* arg)

{ 

  // Wait for any item to be present in queue

  sem_wait(&isNotFullSem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  enter(value);



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not empty

  //  (probably not the best name, but similar to condition)

  sem_post(&isNotEmptySem);

}



void* consume(void* arg)

{

  // Wait for empty space to be present in queue

  sem_wait(&isNotEmptySem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  value = leave();



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not full

  sem_post(&isNotFullSem);

}

answered Nov 25 '18 at 8:40

Paul

3606

I appreciate the feedback but I keep on getting a deadlock for some reason

– James Williams
Nov 25 '18 at 9:50

@JamesWilliams Hmm, I can't see how the code I have above would result in a deadlock. Can you post your updates to your question so I can take a look? Does the deadlock occur only when the program should terminate? You need some handling for that case (i.e. set exit = 1 and post isNotFull and isNotEmpty to make threads return).

– Paul
Nov 25 '18 at 9:56

put my update using your psuedocode as an answer, thanks

– James Williams
Nov 25 '18 at 13:11

add a comment |

Check me on this code -- there may be some issues, but here's what I'd recommend (it's pseudo C)

void* produce(void* arg)

{ 

  // Wait for any item to be present in queue

  sem_wait(&isNotFullSem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  enter(value);



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not empty

  //  (probably not the best name, but similar to condition)

  sem_post(&isNotEmptySem);

}



void* consume(void* arg)

{

  // Wait for empty space to be present in queue

  sem_wait(&isNotEmptySem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  value = leave();



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not full

  sem_post(&isNotFullSem);

}

answered Nov 25 '18 at 8:40

Paul

3606

I appreciate the feedback but I keep on getting a deadlock for some reason

– James Williams
Nov 25 '18 at 9:50

@JamesWilliams Hmm, I can't see how the code I have above would result in a deadlock. Can you post your updates to your question so I can take a look? Does the deadlock occur only when the program should terminate? You need some handling for that case (i.e. set exit = 1 and post isNotFull and isNotEmpty to make threads return).

– Paul
Nov 25 '18 at 9:56

put my update using your psuedocode as an answer, thanks

– James Williams
Nov 25 '18 at 13:11

add a comment |

Check me on this code -- there may be some issues, but here's what I'd recommend (it's pseudo C)

void* produce(void* arg)

{ 

  // Wait for any item to be present in queue

  sem_wait(&isNotFullSem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  enter(value);



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not empty

  //  (probably not the best name, but similar to condition)

  sem_post(&isNotEmptySem);

}



void* consume(void* arg)

{

  // Wait for empty space to be present in queue

  sem_wait(&isNotEmptySem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  value = leave();



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not full

  sem_post(&isNotFullSem);

}

answered Nov 25 '18 at 8:40

Paul

3606

Check me on this code -- there may be some issues, but here's what I'd recommend (it's pseudo C)

void* produce(void* arg)

{ 

  // Wait for any item to be present in queue

  sem_wait(&isNotFullSem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  enter(value);



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not empty

  //  (probably not the best name, but similar to condition)

  sem_post(&isNotEmptySem);

}



void* consume(void* arg)

{

  // Wait for empty space to be present in queue

  sem_wait(&isNotEmptySem);



  // Take "mutex" to protect buffer

  sem_wait(&mutex);



  // Enqueue new value

  value = leave();



  // Release mutex

  sem_post(&mutex);



  // Signal that buffer is not full

  sem_post(&isNotFullSem);

}

answered Nov 25 '18 at 8:40

Paul

3606

answered Nov 25 '18 at 8:40

Paul

3606

answered Nov 25 '18 at 8:40

Paul

3606

answered Nov 25 '18 at 8:40

Paul

3606

I appreciate the feedback but I keep on getting a deadlock for some reason

– James Williams
Nov 25 '18 at 9:50

@JamesWilliams Hmm, I can't see how the code I have above would result in a deadlock. Can you post your updates to your question so I can take a look? Does the deadlock occur only when the program should terminate? You need some handling for that case (i.e. set exit = 1 and post isNotFull and isNotEmpty to make threads return).

– Paul
Nov 25 '18 at 9:56

put my update using your psuedocode as an answer, thanks

– James Williams
Nov 25 '18 at 13:11

add a comment |

I appreciate the feedback but I keep on getting a deadlock for some reason

– James Williams
Nov 25 '18 at 9:50

@JamesWilliams Hmm, I can't see how the code I have above would result in a deadlock. Can you post your updates to your question so I can take a look? Does the deadlock occur only when the program should terminate? You need some handling for that case (i.e. set exit = 1 and post isNotFull and isNotEmpty to make threads return).

– Paul
Nov 25 '18 at 9:56

put my update using your psuedocode as an answer, thanks

– James Williams
Nov 25 '18 at 13:11

I appreciate the feedback but I keep on getting a deadlock for some reason

– James Williams
Nov 25 '18 at 9:50

@JamesWilliams Hmm, I can't see how the code I have above would result in a deadlock. Can you post your updates to your question so I can take a look? Does the deadlock occur only when the program should terminate? You need some handling for that case (i.e. set exit = 1 and post isNotFull and isNotEmpty to make threads return).

– Paul
Nov 25 '18 at 9:56

put my update using your psuedocode as an answer, thanks

– James Williams
Nov 25 '18 at 13:11

add a comment |

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