dnscache --源码笔记

Posted 2020-10-04 张伯雨

//最新版本参见https://github.com/karlseguin/dnscache地址 或者  https://github.com/netroby/dnscache   

package dnscache

// Package dnscache caches DNS lookups

import (

  "net"

  "sync"

  "time"

)

//一个域名可能对应多个ip

type Resolver struct {

  lock sync.RWMutex   //锁--读写锁

  cache map[string][]net.IP  //map中key 为address ，value为域名对应的ip结合集合

}

//参数 refreshRate   设置刷新周期

func New(refreshRate time.Duration) *Resolver {

  resolver := &Resolver {

    cache: make(map[string][]net.IP, 64),

  }

  if refreshRate > 0 {//当刷新周期大于零  开启一个协程来自动解析域名，再次填充cache

    go resolver.autoRefresh(refreshRate)

  }

  return resolver

}

//解析域名对应的ip集合  如果在当前缓存中不存在  ，重新解析并返回

func (r *Resolver) Fetch(address string) ([]net.IP, error) {

  r.lock.RLock()

  ips, exists := r.cache[address]

  r.lock.RUnlock()

  if exists { return ips, nil }

  return r.Lookup(address)

}

//解析域名对应的ip集合中第一个ip

func (r *Resolver) FetchOne(address string) (net.IP, error) {

  ips, err := r.Fetch(address)

  if err != nil || len(ips) == 0 { return nil, err}

  return ips[0], nil

}

//解析域名对应的ip集合第一个ip的String方法

func (r *Resolver) FetchOneString(address string) (string, error) {

  ip, err := r.FetchOne(address)

  if err != nil || ip == nil { return "", err }

  return ip.String(), nil

}

//解析当前cache中所有的域名 并且处于阻塞中2秒  。每一个域名对应一个ip集合 并存储在cache中

func (r *Resolver) Refresh() {

  i := 0

  r.lock.RLock()//获取读锁

  addresses := make([]string, len(r.cache)) //创建一个slice  初始容量为 cache大小  用来存储当前的域名

  for key, _ := range r.cache {

    addresses[i] = key

    i++

  }

  r.lock.RUnlock()

  for _, address := range addresses {

    r.Lookup(address)

    time.Sleep(time.Second * 2)

  }

}

//通过net包 解析域名对应的ip集合

func (r *Resolver) Lookup(address string) ([]net.IP, error) {

  ips, err := net.LookupIP(address)

  if err != nil { return nil, err }

  r.lock.Lock()

  r.cache[address] = ips

  r.lock.Unlock()

  return ips, nil

}

//参数 rate   刷新周期

func (r *Resolver) autoRefresh(rate time.Duration) {

  for {

    time.Sleep(rate) //睡眠rate周期  并且处于阻塞中

    r.Refresh()

  }

}