伪最近最少使用逻辑（Plru）

简介

伪最近最少使用组合逻辑
io.context.state 需由外部逻辑处理。
当需要指定对某表项的访问时，可通过 io.update 接口获取新的状态值。
plru.io.evict.id 指示下一个将被淘汰的块 ID
plru.io.update.id 用于更新最近使用的表项

伪LRU 代码

 val io = new Bundle{
  val context = new Bundle{
    //user -> plru, specify the current state
    val state = Plru.State(entries)
    //user -> plru, allow to specify prefered entries to remove. each bit set mean : "i would prefer that way to not to be selected by PLRU"
    val valids = withEntriesValid generate Bits(entries bits)
  }
  val evict = new Bundle{
    //PLRU -> user, Tells you the least recently used entry for the given context provided above
    val id =  UInt(log2Up(entries) bits)
  }
  val update = new Bundle{
    // user -> PLRU specify which entry the user want to mark as most recently used
    val id = UInt(log2Up(entries) bits)
   // PLRU -> user specfy what should then be the new value of the PLRU status
    val state = Plru.State(entries)
  }
}

缓存中使用示例

val plru = new Area {
   // Define a Mem, to track the state of each set
   val ram = Mem.fill(nSets)(Plru.State(wayCount))
   val write = ram.writePort
   val fromLoad, fromStore = cloneOf(write)
   write.valid := fromLoad.valid || fromStore.valid
   write.payload := fromLoad.valid.mux(fromLoad.payload, fromStore.payload)
}

获取待淘汰路（way）的ID

val replacedWay = plru.io.evict.id

更新最近使用的路

plru.update.id := refillWay