函数逻辑报告 | 
Source Code:block\kyber-iosched.c | 
Create Date:2022-07-27 19:27:29 | 
| Last Modify:2020-03-12 14:18:49 | Copyright©Brick | 
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函数名称:Calculate the histogram bucket with the given percentile rank, or -1 if there* aren't enough samples yet.
函数原型:static int calculate_percentile(struct kyber_queue_data *kqd, unsigned int sched_domain, unsigned int type, unsigned int percentile)
返回类型:int
参数:
| 类型 | 参数 | 名称 | 
|---|---|---|
| struct kyber_queue_data * | kqd | |
| unsigned int | sched_domain | |
| unsigned int | type | |
| unsigned int | percentile | 
| 229 | buckets等于latency_buckets[sched_domain][type] | 
| 230 | samples等于0 | 
| 235 | 如果非samples则返回:负1 | 
| 242 | 如果非latency_timeout[sched_domain]则latency_timeout[sched_domain]等于两数取大(jiffies + HZ, 1UL) | 
| 246 | 返回:负1 | 
| 248 | latency_timeout[sched_domain]等于0 | 
| 250 | percentile_samples等于DIV_ROUND_UP(samples * percentile, 100) | 
| 251 | 以bucket小于KYBER_LATENCY_BUCKETS减1循环 | 
| 252 | 如果buckets[bucket]大于等于percentile_samples则退出 | 
| 254 | percentile_samples减等于buckets[bucket] | 
| 256 | memset(buckets, 0, latency_buckets[sched_domain][type]的长度) | 
| 258 | trace_kyber_latency(q, kyber_domain_names[sched_domain], kyber_latency_type_names[type], percentile, bucket + 1, 1 << KYBER_LATENCY_SHIFT, samples) | 
| 262 | 返回:bucket | 
| 名称 | 描述 | 
|---|---|
| kyber_timer_fn | 
| 源代码转换工具 开放的插件接口  | X | 
|---|---|
| 支持:c/c++/esqlc/java Oracle/Informix/Mysql 插件可实现:逻辑报告 代码生成和批量转换代码  |