pthread Mutexes

[newNode]

hey guys, i am trying to print "begin" then 11, i've been trying many different ways but I still get race condition, can anyone help? thanks.

#include <pthread.h>
#include <semaphore.h>
#include <stdio.h>
   
pthread_mutex_t  m1;  //declares m1 to be a mutex
int  x;
   
//------------------------------------------------------
void *thread1(void *arg)  {
     //pthread_mutex_lock(&m1);   
     pthread_mutex_unlock(&m1);

     printf("begin");  
     x=x+1;
     printf("\n%d \n", x);
    
     }
//------------------------------------------------------
   
//------------------------------------------------------
void *thread2(void *arg)  {
     pthread_mutex_lock(&m1);
     
     x=10;
     pthread_mutex_unlock(&m1);
     }
//------------------------------------------------------

//------------------------------------------------------
int main (void)  {
   pthread_t TId1, TId2;
   int status;
      
   pthread_mutex_init(&m1, NULL);  //MInitialize(m1)
   x=0;
      
   //create threads
   pthread_create(&TId1, NULL, thread1, NULL);
   pthread_create(&TId2, NULL, thread2, NULL);
      
   //wait for joinable threads to complete
   pthread_join(TId1, (void **)&status);
   pthread_join(TId2, (void **)&status);
      
  
   
   printf("push Enter when done looking\n"); getchar();
   pthread_exit(NULL);
}
//------------------------------------------------------ 

[helios]

I think initializing mutexes in pthreads is a little more complicated than that.

pthread_mutexattr_t attr;
pthread_mutexattr_init(&attr);
pthread_mutexattr_settype(&attr,PTHREAD_MUTEX_RECURSIVE);
pthread_mutex_init(&mutex,&attr);
pthread_mutexattr_destroy(&attr);

[newNode]

this code is from my professor. It does work. The only thing is I need to make sure it prints out
begin
11

this is his question:

Consider:
   
do par
thread1:
do
print "begin";
x:=x+1;
print x;
end;  // thread1
   
thread2:
do
x:=10;
end;  // thread2
end par;
   
//Insert MInitialize(m), MRequest(m)'s, and MRelease(m)'s for a mutex m into the above code so that it is guaranteed to print:
   
begin
11
   
//Don't otherwise change the existing code - only insert the above function calls.  Hint: A thread may release a mutex without having requesting it.  You don't need to translate and run this. 

I've spent so many hours and asked so many people.. lol.. someone know this please help, thanks!

[helios]

Ah, there it is.
Line 11: Don't unlock a mutex in a thread when that thread hasn't locked it.
As a rule of thumb, abs((number of locks)-(number of unlocks))<=(how many problems you have)
Just add a set of lock-unlock around line 14 and you should be fine.

[newNode]

ok thanks a lot Helios. You mean

do par
thread1:
do
print "begin";
x:=x+1;
print x;
end;  // thread1
   
thread2:
do
MRequest(m);
x:=10;
MRelease(m);
end;  // thread2
end par;

? because line 14, that's end;