关于UNIX高级环境编程2 中第11章线程死锁的例程代码，看不懂 望高手指点
#include <stdlib.h>
#include <pthread.h>

#define NHASH 29
#define HASH(fp) (((unsigned long)fp)%NHASH) // 请问这条语句的作用？
struct foo *fh[NHASH];

pthread_mutex_t hashlock = PTHREAD_MUTEX_INITIALIZER;

struct foo {
  int f_count;
  pthread_mutex_t f_lock;
  struct foo *f_next; /* protected by hashlock */
  int f_id;
  /* ... more stuff here ... */
};

struct foo *
foo_alloc(void) /* allocate the object */
{
  struct foo *fp;
  int idx;

  if ((fp = malloc(sizeof(struct foo))) != NULL) {
  fp->f_count = 1;
  if (pthread_mutex_init(&fp->f_lock, NULL) != 0) {
  free(fp);
  return(NULL);
  }
  idx = HASH(fp);
  pthread_mutex_lock(&hashlock);
  fp->f_next = fh[idx]; // 请问这条语句的作用？
  fh[idx] = fp->f_next; // 请问这条语句的作用？
  pthread_mutex_lock(&fp->f_lock);
  pthread_mutex_unlock(&hashlock);
  /* ... continue initialization ... */
  pthread_mutex_unlock(&fp->f_lock);
  }
  return(fp);
}

void
foo_hold(struct foo *fp) /* add a reference to the object */
{
  pthread_mutex_lock(&fp->f_lock);
  fp->f_count++;
  pthread_mutex_unlock(&fp->f_lock);
}

struct foo *
foo_find(int id) /* find an existing object */
{
  struct foo *fp;
  int idx;

  idx = HASH(fp);

  pthread_mutex_lock(&hashlock);
  for (fp = fh[idx]; fp != NULL; fp = fp->f_next) {
  if (fp->f_id == id) {
  foo_hold(fp);
  break;
  }
  }
  pthread_mutex_unlock(&hashlock);
  return(fp);
}

void
foo_rele(struct foo *fp) /* release a reference to the object */
{
  struct foo *tfp;
  int idx;

  pthread_mutex_lock(&fp->f_lock);
  if (fp->f_count == 1) { /* last reference */ // 矛盾1
  pthread_mutex_unlock(&fp->f_lock);
  pthread_mutex_lock(&hashlock);
  pthread_mutex_lock(&fp->f_lock);
  /* need to recheck the condition */
  if (fp->f_count != 1) { //矛盾2
  fp->f_count--;
  pthread_mutex_unlock(&fp->f_lock);
  pthread_mutex_unlock(&hashlock);
  return;
  }
  /* remove from list */
  idx = HASH(fp);
  tfp = fh[idx];
  if (tfp == fp) {
  fh[idx] = fp->f_next;
  } else {
  while (tfp->f_next != fp)
  tfp = tfp->f_next;
  tfp->f_next = fp->f_next;
  }
  pthread_mutex_unlock(&hashlock);
  pthread_mutex_unlock(&fp->f_lock);
  pthread_mutex_destroy(&fp->f_lock);
  free(fp);
  } else {
  fp->f_count--;
  pthread_mutex_unlock(&fp->f_lock);
  }
}

请教大家一个问题 这是UNXI高级环境编程第11章 讲线程死锁 的例程  
有几个地方看不明白 想请问大家帮忙解释下 看不明白的地方已经注注释了 ，还有在foo_rele函数中 注释的矛盾1和矛盾2 两条语句 不是相互矛盾吗？ 既然条件是 if(fp->f_count==1),那在它的嵌套里面为什么还有一个if(fp->f_count!=1),既然等于1才执行后面语句，那后面的！=1