线程锁模拟交叉点

我有去海峡的功能,左转,右转.
现在,我有一个接近十字路口的功能.这会产生随机方向并转向.每个线程共享接近的交叉点.

我为所有方向的所有汽车定义了所有锁.

采取进入海峡的功能.它有一个switch语句,可以打印当时汽车正在做什么.现在,我只是不确定锁定此功能的内容.如果汽车指向北方的方向,我将锁定东西方向,同时指向南方向北的汽车？

这是我的锁只调用一个锁定或解锁的功能

#define NUMCARS 30

#define lock_NW(CAR) lock(CAR,NW_mutex)
#define lock_NE(CAR) lock(CAR,NE_mutex)
#define lock_SW(CAR) lock(CAR,SW_mutex)
#define lock_SE(CAR) lock(CAR,SE_mutex)

#define unlock_NW(CAR) unlock(CAR,NW_mutex)
#define unlock_NE(CAR) unlock(CAR,NE_mutex)
#define unlock_SW(CAR) unlock(CAR,SW_mutex)
#define unlock_SE(CAR) unlock(CAR,SE_mutex)

这里是主要的

int main(int argc,char **argv){
/* Initial variables*/
int index,tid;
unsigned int carids[NUMCARS];
pthread_t carthreads[NUMCARS];

/* Start up a thread for each car*/ 
for(index = 0; index <NUMCARS; index++){
carids[index] = index;
tid = pthread_create(&carthreads[index],NULL,approachintersection,(void*)&carids[index]);
}

/* Wait for every car thread to finish */
for(index = 0; index <NUMCARS; index++){
pthread_join(carthreads[index],NULL);
}
printf("Donen");
return 1;
}

这是一个即将到来的交叉路口,它将功能称为海峡

static void * approachintersection(void* arg){
unsigned int * carnumberptr;
unsigned int carnumber;
orientation_t cardir = (orientation_t)random()%4;
unsigned long turn = random()%3;

carnumberptr = (unsigned int*) arg;
carnumber = (unsigned int) *carnumberptr;

if(turn==LEFT){
turnleft(cardir,carnumber);
} else if(turn==RIGHT){
turnright(cardir,carnumber);
} else {//straight
gostraight(cardir,carnumber);
}

return (void*)carnumberptr;
}

现在,这是我要锁定适当方向的海峡功能.

/*
  cardirection - The direction the car is pointing.  If it is pointing NORTH,it is starting from the South-Eastern corner of the intersection
  and "going straight" means it wants to move SOUTH to NORTH.

  valid options: NORTH,SOUTH,EAST,WEST

 carnumber -    The car identifier
*/


static void gostraight(orientation_t cardirection,unsigned int carnumber){

switch(cardirection){
case NORTH:
printf("Car %d,Moving South-Northn",carnumber);
break;
case SOUTH:
printf("Car %d,Moving North-Southn",carnumber);
break;
case EAST:
printf("Car %d,Moving West-Eastn",carnumber);
break;
case WEST:
printf("Car %d,Moving East-Westn",carnumber);
break;
}
}

因此,如果接近的汽车从南方指向北方,那么汽车将是SE汽车,我会用lock_SE(CAR)锁定箱子东,西打印功能？防止其他线程进入并打印？所以我会锁定解锁打印语句？

或者我会锁定整个switch语句？

**编辑：这会是这样做的吗？ **

static void turnleft(orientation_t cardirection,unsigned int carnumber){

int CAR;
CAR = carnumber;


  switch(cardirection){
  case NORTH:
  lock_SE(CAR)
  printf("Car %d,Moving South-Westn",carnumber);
  unlock_SE(CAR)
  break;
  case SOUTH:
  lock_NW(CAR)
  printf("Car %d,Moving North-Eastn",carnumber);
  unlock_NW(CAR)
  break;
  case EAST:
  lock_SW(CAR)
  printf("Car %d,Moving West-Northn",carnumber);
  unlock_SW(CAR)
  break;
  case WEST:
  lock_NE(CAR)
  printf("Car %d,Moving East-Southn",carnumber);
  unlock_NE(CAR)
  break;
  }

}