目标

• 维护一段win时间内的三个best最大值或者最小值，以最大值为例
• 值大小上，1st best > 2nd best > 3rd best
• 采集时间，1st best < 2nd best < 3rd best
  
  注：win时间，表示一个窗口时间，或者说一段时间

代码分析

以最大值为例进行分析

基本结构体

一个样本，包含一个时间值和一个大小，其中时间值没有单位，由调用者决定，只要保证和win时间的单位一致即可，所以可以是秒，可以是10s

/* A single data point for our parameterized min-max tracker */
struct minmax_sample {
	u32	t;	/* time measurement was taken */
	u32	v;	/* value measured */
};

状态跟踪结构体，三个样本

/* State for the parameterized min-max tracker */
struct minmax {
	struct minmax_sample s[3];
};

重置

static inline u32 minmax_reset(struct minmax *m, u32 t, u32 meas)
{
	struct minmax_sample val = { .t = t, .v = meas };

	m->s[2] = m->s[1] = m->s[0] = val;
	return m->s[0].v;
}

将三个best值全部设置为传入的时间值和样本值。

更新最大值

/* Check if new measurement updates the 1st, 2nd or 3rd choice max. */
u32 minmax_running_max(struct minmax *m, u32 win, u32 t, u32 meas)
{
	struct minmax_sample val = { .t = t, .v = meas };

	if (unlikely(val.v >= m->s[0].v) ||	  /* found new max? */
	    unlikely(val.t - m->s[2].t > win))	  /* nothing left in window? */
		return minmax_reset(m, t, meas);  /* forget earlier samples */

	if (unlikely(val.v >= m->s[1].v))
		m->s[2] = m->s[1] = val;
	else if (unlikely(val.v >= m->s[2].v))
		m->s[2] = val;

	return minmax_subwin_update(m, win, &val);
}

用流程图表示：

刷新时间

/* As time advances, update the 1st, 2nd, and 3rd choices. */
static u32 minmax_subwin_update(struct minmax *m, u32 win,
				const struct minmax_sample *val)
{
	u32 dt = val->t - m->s[0].t;

	if (unlikely(dt > win)) {
		/*
		 * Passed entire window without a new val so make 2nd
		 * choice the new val & 3rd choice the new 2nd choice.
		 * we may have to iterate this since our 2nd choice
		 * may also be outside the window (we checked on entry
		 * that the third choice was in the window).
		 */
		m->s[0] = m->s[1];
		m->s[1] = m->s[2];
		m->s[2] = *val;
		if (unlikely(val->t - m->s[0].t > win)) {
			m->s[0] = m->s[1];
			m->s[1] = m->s[2];
			m->s[2] = *val;
		}
	} else if (unlikely(m->s[1].t == m->s[0].t) && dt > win/4) {
		/*
		 * We've passed a quarter of the window without a new val
		 * so take a 2nd choice from the 2nd quarter of the window.
		 */
		m->s[2] = m->s[1] = *val;
	} else if (unlikely(m->s[2].t == m->s[1].t) && dt > win/2) {
		/*
		 * We've passed half the window without finding a new val
		 * so take a 3rd choice from the last half of the window
		 */
		m->s[2] = *val;
	}
	return m->s[0].v;
}

用流程图表示：

有几个问题

• 为什么更新两次全部更新之后不再比较3rd与样本值？
• 因为在更新最大值的第一步流程中就比较过新样本和3rdbest的时间戳，所以这里保证了3rdbest的时间戳和样本的时间戳差值肯定在win时间内
• 如果2nd的值未设置过，则需要等待win的0.25倍时间
• 如果3rd的值未设置过，则需要等待win的0.5倍时间

总结

• win_minmax维护了win时间内的最优值
• 当1st best已经不属于当前时间的win窗口时间内时，会由2nd best顶上，甚至3rd best顶上

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