操作系统：内存分配

Question

问题图片：

鉴于上述问题，我有几个问题。首先，proc0代替了一个比它更大的洞，剩下的空间会发生什么？

例如在Q1 i中：引入Proc0之后它是a. 10->15->15->25->30。让15替换20，theres 5离开，那么它会发生什么，我将如何描述它发生了什么？它会是一个。 10-> 5-> 15-> 15-> 25-> 30或a。 10→15-＆GT; 5→15-＆GT; 25-＆GT; 30

Answer 1

给定，proc

的大小

sz（proc 0） - ＆gt; 15 sz（proc 1） - ＆gt; 5

根据我的理解，Free List (10 -> 20 -> 15 -> 25 -> 30)的变化描述如下：

  

Q1：First Fit（自由列表的大小将开始减少）

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

假设在执行First Fit分配之前FreeList再次相同(10 -> 20 -> 15 -> 25 -> 30)：

  

Q2：Next Fit（自由列表的大小将开始减少）

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