# 2021q1 Homework3 (fibdrv) contributed by < `Tcc0403` > [作業要求](https://hackmd.io/@sysprog/linux2022-fibdrv#-%E4%BD%9C%E6%A5%AD%E8%A6%81%E6%B1%82) ## 自我檢查清單 - [ ] 研讀上述 ==Linux 效能分析的提示== 描述,在自己的實體電腦運作 GNU/Linux,做好必要的設定和準備工作 $\to$ 從中也該理解為何不希望在虛擬機器中進行實驗; - [ ] 研讀上述費氏數列相關材料 (包含論文),摘錄關鍵手法,並思考 [clz / ctz](https://en.wikipedia.org/wiki/Find_first_set) 一類的指令對 Fibonacci 數運算的幫助。請列出關鍵程式碼並解說 - [ ] 複習 C 語言 [數值系統](https://hackmd.io/@sysprog/c-numerics) 和 [bitwise operation](https://hackmd.io/@sysprog/c-bitwise),思考 Fibonacci 數快速計算演算法的實作中如何減少乘法運算的成本; - [ ] 研讀 [KYG-yaya573142 的報告](https://hackmd.io/@KYWeng/rkGdultSU),指出針對大數運算,有哪些加速運算和縮減記憶體操作成本的舉措? - [ ] `lsmod` 的輸出結果有一欄名為 `Used by`,這是 "each module's use count and a list of referring modules",但如何實作出來呢?模組間的相依性和實際使用次數 ([reference counting](https://en.wikipedia.org/wiki/Reference_counting)) 在 Linux 核心如何追蹤呢? > 搭配閱讀 [The Linux driver implementer’s API guide » Driver Basics](https://www.kernel.org/doc/html/latest/driver-api/basics.html) - [ ] 注意到 `fibdrv.c` 存在著 `DEFINE_MUTEX`, `mutex_trylock`, `mutex_init`, `mutex_unlock`, `mutex_destroy` 等字樣,什麼場景中會需要呢?撰寫多執行緒的 userspace 程式來測試,觀察 Linux 核心模組若沒用到 mutex,到底會發生什麼問題。嘗試撰寫使用 [POSIX Thread](https://en.wikipedia.org/wiki/POSIX_Threads) 的程式碼來確認。 ## 實驗環境 ```shell $ lscpu Architecture: x86_64 CPU op-mode(s): 32-bit, 64-bit Byte Order: Little Endian Address sizes: 48 bits physical, 48 bits virtual CPU(s): 12 On-line CPU(s) list: 0-11 Thread(s) per core: 2 Core(s) per socket: 6 Socket(s): 1 NUMA node(s): 1 Vendor ID: AuthenticAMD CPU family: 25 Model: 33 Model name: AMD Ryzen 5 5600X 6-Core Processor Stepping: 0 Frequency boost: enabled CPU MHz: 2200.000 CPU max MHz: 4650.2920 CPU min MHz: 2200.0000 BogoMIPS: 7385.92 Virtualization: AMD-V L1d cache: 192 KiB L1i cache: 192 KiB L2 cache: 3 MiB L3 cache: 32 MiB NUMA node0 CPU(s): 0-11 ```
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