OS：內存分配

Question

圖片問題：

鑑於以上問題我有幾個問題。首先，proc0代替了一個比它大的孔，剩下的空間會發生什麼？

例如在Q1我：在引入Proc0後它是a. 10->15->15->25->30。有15個替換20個，剩下的5個，所以它會發生什麼，我將如何描述它會發生什麼？它會是一個嗎？ 10→5→15→15→25→30或a。 10-> 15-> 5-> 15-> 25-> 30？

Answer 1

考慮，該proc

SZ的尺寸（PROC 0） - > 15 SZ（PROC 1） - > 5

從什麼我能理解，在Free List (10 -> 20 -> 15 -> 25 -> 30)的變化說明如下：

Q1：第一飛度（自由列表的大小將開始下降）

When proc 0 (size=15) is brought to the list, 
Free List converts to (10 -> 5 -> 15 -> 25 -> 30) 
// the freelist size would decrease wherever the first biggest hole is found, 
// so, hole of size 20 is replaced by a hole of size 5 to allocate memory to proc 0 

When proc 1 (size=5) is brought to the list, 
Free List converts to (5 -> 5 -> 15 -> 25 -> 30) 
// the freelist size would decrease wherever the first biggest hole is found, 
// so, hole of size 10 is replaced by a hole of size 5 to allocate memory to proc 1 

假設空閒列表再次是一樣的(10 -> 20 -> 15 -> 25 -> 30)執行第一嵌入分配之前：

Q2：接下來裝（自由列表的大小將開始下降）

When proc 0 (size=15) is brought to the list, 
Free List converts to (10 -> 5 -> 15 -> 25 -> 30) 
// the freelist size would decrease wherever the next biggest hole is found, 
// so, hole of size 20 is replaced by a hole of size 5 to allocate memory to proc 0 

When proc 1 (size=5) is brought to the list, 
Free List converts to (10 -> 0 -> 15 -> 25 -> 30), or better 
Free List converts to (10 -> 15 -> 25 -> 30) // the size of the freelist decreases. 
// the freelist size would decrease wherever the next biggest hole is found, 
// so, hole of size 5 is replaced by a hole of size 0 
// (or, rather no hole left, so list becomes continuous) to allocate memory to proc 1