考虑具有 4 个组和总共 8 个高速缓存块 (0-7) 的 2 路组关联高速缓存存储器和具有 128 个块 (0-127) 的主存储器。如果 LRU 策略用于缓存块替换，那么在以下内存块引用序列之后，哪些内存块将出现在缓存中。假设最初缓存没有来自当前作业的任何内存块?
0 5 3 9 7 0 16 55
(A) 0 3 5 7 16 55
(B) 0 3 5 7 9 16 55
(C) 0 5 7 9 16 55
(D) 3 5 7 9 16 55答案： (C)
解释：
2-way set associative cache memory, .i.e K = 2.
No of sets is given as 4, i.e. S = 4 ( numbered 0 - 3 )
No of blocks in cache memory is given as 8, i.e. N =8 ( numbered from 0 -7)
Each set in cache memory contains 2 blocks.
The number of blocks in the main memory is 128, i.e M = 128. ( numbered from 0 -127)
A referred block numbered X of the main memory is placed in the
set numbered ( X mod S ) of the the cache memory. In that set, the
block can be placed at any location, but if the set has already become
full, then the current referred block of the main memory should replace
a block in that set according to some replacement policy. Here
the replacement policy is LRU ( i.e. Least Recently Used block should
be replaced with currently referred block).
X ( Referred block no ) and
the corresponding Set values are as follows:
X-->set no ( X mod 4 )
0--->0 ( block 0 is placed in set 0, set 0 has 2 empty block locations,
block 0 is placed in any one of them )
5--->1 ( block 5 is placed in set 1, set 1 has 2 empty block locations,
block 5 is placed in any one of them )
3--->3 ( block 3 is placed in set 3, set 3 has 2 empty block locations,
block 3 is placed in any one of them )
9--->1 ( block 9 is placed in set 1, set 1 has currently 1 empty block location,
block 9 is placed in that, now set 1 is full, and block 5 is the
least recently used block )
7--->3 ( block 7 is placed in set 3, set 3 has 1 empty block location,
block 7 is placed in that, set 3 is full now,
and block 3 is the least recently used block)
0--->block 0 is referred again, and it is present in the cache memory in set 0,
so no need to put again this block into the cache memory.
16--->0 ( block 16 is placed in set 0, set 0 has 1 empty block location,
block 0 is placed in that, set 0 is full now, and block 0 is the LRU one)
55--->3 ( block 55 should be placed in set 3, but set 3 is full with block 3 and 7,
hence need to replace one block with block 55, as block 3 is the least
recently used block in the set 3, it is replaced with block 55.
因此，高速缓存内存中的主内存块为： 0, 5, 7, 9, 16, 55 。
(注意：缓存中不存在块 3，它被块 55 替换)
阅读以下文章以了解与上述问题相关的更多信息：
高速缓存存储器
缓存组织 |介绍这个问题的测验