内核延时函数

1) msleep:实现毫秒级的延时，该延时保证至少延时所设置的延时时间，不会提前超时返回，会让出CPU

void msleep(unsigned int msecs)
{
	unsigned long timeout = msecs_to_jiffies(msecs) + 1;

	while (timeout)
		timeout = schedule_timeout_uninterruptible(timeout);
}

为什么在转换成jiffies时要+1呢？前边我们讲到，该延时要至少保证延时转换的jiffies时间，一个jiffies为10毫秒，比如我们可以查10个数表示一个jiffies，在数到5时调用了msleep，那么显然我们不能在此jiffies到时时返回，违反了至少延时设置的jiffies的原则，因此转换成jiffies+1是比较合适的，内核中也特意做了解释。

unsigned long msecs_to_jiffies(const unsigned int m)
{
	/*
	 * Negative value, means infinite timeout:
	 */
	if ((int)m < 0)
		return MAX_JIFFY_OFFSET;
。
。
。
}
/*
 * Change timeval to jiffies, trying to avoid the
 * most obvious overflows..
 *
 * And some not so obvious.
 *
 * Note that we don't want to return LONG_MAX, because
 * for various timeout reasons we often end up having
 * to wait "jiffies+1" in order to guarantee that we wait
 * at _least_ "jiffies" - so "jiffies+1" had better still
 * be positive.
 */
#define MAX_JIFFY_OFFSET ((LONG_MAX >> 1)-1)

2)msleep_interruptible:毫秒级延时，该延时函数有可能被信号打断提前超时，返回剩余的时间，会让出CPU

unsigned long msleep_interruptible(unsigned int msecs)
{
	unsigned long timeout = msecs_to_jiffies(msecs) + 1;

	while (timeout && !signal_pending(current))
		timeout = schedule_timeout_interruptible(timeout);
	return jiffies_to_msecs(timeout);
}

3)ssleep:秒级延时，通过调用msleep实现,会让出CPU

static inline void ssleep(unsigned int seconds)
{
	msleep(seconds * 1000);
}

4)usleep_range:该延时函数实现微秒级延时，特别之处在于其可以设定一个超时范围，通过看源代码可以发现此函数设置任务状态为ASK_UNINTERRUPTIBLE，即该延时至少可以保证延时min微秒而不被打断。会让出CPU

/**
 * usleep_range - Drop in replacement for udelay where wakeup is flexible
 * @min: Minimum time in usecs to sleep
 * @max: Maximum time in usecs to sleep
 */
void usleep_range(unsigned long min, unsigned long max)
{
	__set_current_state(TASK_UNINTERRUPTIBLE);
	do_usleep_range(min, max);
}

5)ndelay:纳秒级延时,不会让出CPU

static inline void ndelay(unsigned long x)
{
	udelay(DIV_ROUND_UP(x, 1000));
}

6)udelay:微秒延时，不会让出CPU

/*
 * division by multiplication: you don't have to worry about
 * loss of precision.
 *
 * Use only for very small delays ( < 2 msec).  Should probably use a
 * lookup table, really, as the multiplications take much too long with
 * short delays.  This is a "reasonable" implementation, though (and the
 * first constant multiplications gets optimized away if the delay is
 * a constant)
 */
#define __udelay(n)		arm_delay_ops.udelay(n)
#define __const_udelay(n)	arm_delay_ops.const_udelay(n)

#define udelay(n)							\
	(__builtin_constant_p(n) ?					\
	  ((n) > (MAX_UDELAY_MS * 1000) ? __bad_udelay() :		\
			__const_udelay((n) * UDELAY_MULT)) :		\
	  __udelay(n))

关于__builtin_constant_p(x)，准确的定义应该是：如果x的值在编译时能确定，那么该函数返回值为1。

对于arm_delay_ops来讲，内核有一套默认的回调函数

/*
 * Default to the loop-based delay implementation.
 */
struct arm_delay_ops arm_delay_ops = {
	.delay		= __loop_delay,
	.const_udelay	= __loop_const_udelay,
	.udelay		= __loop_udelay,
};

但是大部分厂商一般都会注册自己的timer，来提供延时使用，具体可参考arch/arm/lib/delay.c实现

/*
 * Default to the loop-based delay implementation.
 */
struct arm_delay_ops arm_delay_ops = {
	.delay		= __loop_delay,
	.const_udelay	= __loop_const_udelay,
	.udelay		= __loop_udelay,
};

#ifdef ARCH_HAS_READ_CURRENT_TIMER
static void __timer_delay(unsigned long cycles)
{
	cycles_t start = get_cycles();

	while ((get_cycles() - start) < cycles)
		cpu_relax();
}

static void __timer_const_udelay(unsigned long xloops)
{
	unsigned long long loops = xloops;
	loops *= loops_per_jiffy;
	__timer_delay(loops >> UDELAY_SHIFT);
}

static void __timer_udelay(unsigned long usecs)
{
	__timer_const_udelay(usecs * UDELAY_MULT);
}

void __init init_current_timer_delay(unsigned long freq)
{
	pr_info("Switching to timer-based delay loop\n");
	lpj_fine			= freq / HZ;//一个jiffy定时器跳变的值
	loops_per_jiffy			= lpj_fine;
	arm_delay_ops.delay		= __timer_delay;
	arm_delay_ops.const_udelay	= __timer_const_udelay;
	arm_delay_ops.udelay		= __timer_udelay;
}

unsigned long __cpuinit calibrate_delay_is_known(void)
{
	return lpj_fine;
}
#endif

7)mdelay:毫秒级延时，ndelay的1000倍，不会让出CPU

#ifndef mdelay
#define mdelay(n) (\
	(__builtin_constant_p(n) && (n)<=MAX_UDELAY_MS) ? udelay((n)*1000) : \
	({unsigned long __ms=(n); while (__ms--) udelay(1000);}))
#endif

 

关于延时函数会不会让出CPU，使用时需要注意，一般对延时要求特别精确，使用不让出CPU的延时函数；对延时要求不是特别精确的，可以使用让出CPU的延时函数，为了保证延时时系统不会进入睡眠，通常在使用会让出CPU的延时前要加上wakelock锁来阻止睡眠。