# Cache

## 觀念

* Cache Size: C = S x E x B data bytes
S = 2s sets
E = 2e lines per set
B = sb bytes per cache block (data)

## 計算


Cache Size: C = S x E x B data bytes

已知：
* L1D-Cache 8 way x6
* Cache Size = 32KBytes
* 1 line = 64byte = 8word x 64bit作業系統
* disk = 1TB

計算：
32KBytes = sets x <font color=red>64B line</font> x <font color=red>8 way</font>
Cache Size / line Size = blocks
32KB / 64B = <font color=red>512b blocks</font>
512b blocks / 8 way = <font color=red>64b sets</font>

Disk = 1TB
2^10 x 2^10 x 2^10 x 2^10 = 2^40B -> <font color=red>全部40bit</font>

| 全部 | tag | index | line offset |
| -------- | -------- | -------- | -------- |
| 40bit | ? | <font color=red>sets</font> 64=2^6 -> <font color=red>6bit</font> | <font color=red>line size</font> 64=2^6 -> <font color=red>6bit</font>|

tag -> 40 - 6 - 6 = <font color=red>28</font>

| vaild | tag | the data = line offset |
| -------- | -------- | -------- |
| 2b | 28b | 6Bx8=512b |

2 + 28 + 512 = 542b

<font color=red>A: L1D-Cache = 542b x 512b blocks x 8快取數量</font>