推 BaaaSwin:謝謝你!!!!我研究看看 11/17 19:30
※ 引述《BaaaSwin (codown)》之銘言：
: 1.有三種RAW,WAR,WAW
: RAW(true dependency): it arises when the next instruction tries to read a
: source before the previous instruction writes to it.
: so, the next instruction gets the old value incorrectly.
下一個instruction嘗試讀取上一個instruction正在寫入的source
導致下一個instruction讀取到舊的值的情形 稱為RAW
: WAR:arises when the next instruction writes to a destination before the
: previous instruction reads it. in the case, the previous instruction
: gets a new value incorrectly.
WAR則是當下一個instruction嘗試在上一個instruction讀取資料之前就在讀取的source
寫入新的資料 導致上個instruction讀取到不正確的值
: WAW:it is situation when the next instruction tries to write to a destination
: done in the wrong order.
而WAW則跟WAR很像 只是改成下一個instruction在上一個instruction寫資料前
就在同個位置上先把資料寫進去了 而上一個instruction後來才寫入資料導致的錯誤
: 2.WAW是否是output dependency,WAR是anti dependency??
yes , yes
: 都可以用register renaming去解決??
yes
: 那什麼又是name dependency呢??
name dependency = anti dependency
指多個instruction用到同樣的register來存資料
但是彼此之間並沒有資料互相流通使用
: 3.Given four instruction, how many unique comparisons(between register sources
: and destinations) are necessary to find all of the RAW,WAR,and WAW dependences.
: Answer for the case of four instructions, and then derive a general equation
: for N instructions. Assume that all instructions have one register destination
: and two register sources.
給四個instruction 問最少要比較register source和destination多少次
才能找到所有的RAW WAR WAW dependence
先回答有四個instruction的情況 再導出一個計算這個問題的公式
希望能回答到你的問題~
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