cpu cache中LRU演算法所需要的位元

標籤：cpu   緩衝   演算法   

參考：algorithm LRU, how many bits needed for implement this algorithm?

問題：在cpu緩衝中使用的LRU替換演算法需要多少位呢？

解決方案：
對於n路相連的緩衝來說，LRU每個緩衝塊需要的位元為log2(n)，那麼每個set需要的位元就為n*log2(n)。

（原文：Assuming you mean a 4-way set-associative cache:
A “perfect” LRU would essentially be assigning each line an exact index in the order of usage. You can also think of that as an “age”. So each of the 4 elements would require an index of 2 bits (since we need to count 4 distinct ages) stating its location in the LRU order - this means 2 bits * 4 ways, per each set of the cache.
In the general case of n ways, you’d need log2(n) bits per line, or n*log2(n) bits per set.

By the way, there are cheaper ways to reach an almost-LRU behavior, see for e.g. Pseudo LRU which would only require 3 bits for the entire set in your case (or in general: #ways - 1)）

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cpu cache中LRU演算法所需要的位元