计算系统调用答案

【问题标题】：Counting system calls计算系统调用
【发布时间】：2011-09-20 01:00:15
【问题描述】：

我有这个 .c 文件来计算 linux 调用的系统调用。这些只是主要功能。我还需要做一些其他的事情，比如创建一个数组

unsigned long syscall_counts[345];

然后在另一个包含一些程序集的文件中，我使用以下命令增加了数组：

incl syscall_counts(,%eax,4)

// This function is called each time the application calls read(). It starts the      process of
// accumulating data to fill the application buffer. Return a pointer representing the current
// item. Return NULL if there are no more items.
//
static void *counter_seq_start(struct seq_file *s, loff_t *record_number)
{
  if (*record_number > 347)
  return NULL;
return (void*)s;
}


// This function is called to compute the next record in the sequence given a pointer to the
// current record (in bookmark). It returns a pointer to the new record (essentially, an updated
// bookmark) and updates *record_number appropriately. Return NULL if there are no more items.
//
static void *counter_seq_next(struct seq_file *s, void *bookmark, loff_t *record_number)
{
   unsigned long *temp_b =(unsigned long*) bookmark;
   (*temp_b)++;
   if (*temp_b > 345)
  return NULL;
   return (void*)temp_b;
}


// This function is called whenever an application buffer is filled (or when start or next
// returns NULL. It can be used to undo any special preparations done in start (such as
// deallocating auxillary memory that was allocated in start. In simple cases, you often do not
// need to do anything in this function.
//
static void  counter_seq_stop(struct seq_file *s, void *bookmark)
{

}


  // This function is called after next to actually compute the output. It can use various seq_...
 // printing functions (such as seq_printf) to format the output. It returns 0 if successful or a
 // negative value if it fails.
 //
 static int counter_seq_show(struct seq_file *s, void *bookmark)
 {
   loff_t *bpos = (loff_t *) bookmark;

   seq_printf(s, "value: %Ld\n", *bpos);

   return 0;
 }


 // Define the only file handling function we need.
 static int counter_open(struct inode *inode, struct file *file)
 {
    return seq_open(file, &counter_seq_ops);
 }

我的输出很奇怪：

有人知道问题出在哪里吗？

【问题讨论】：

^@ 看起来像是在打印一个值为 0 的字符。不过不知道为什么会这样。

标签： c linux security

【解决方案1】：

你不觉得吗：

static int counter_seq_show(struct seq_file *s, void *bookmark) {  
    unsigned long *bpos = (unsigned long *) bookmark;  
    seq_printf(s, "value: %Ld\n", *bpos);  
    return 0;
}

甚至

static int counter_seq_show(struct seq_file *s, void *bookmark) {  
    seq_printf(s, "value: %lu\n", *((unsigned long *)bpos));  
    return 0;
}

我还没有完全理解您的程序，但我看到了您投射“书签”的两种不同方式。在一个函数中，您将其转换为“unsigned long *”，而在另一个函数中，您将其转换为“loff_t *”（long int）。理想情况下，它们应该是相同的，但是出于某种原因，您是否这样做？

HTH

【讨论】：