Skip to content
GitLab
Commit 157d169a authored by 银角's avatar 银角
Browse files

Update README.md

parent 0f7d452e
master
No related merge requests found
Hide whitespace changes
Inline Side-by-side
Showing
with 349 additions and 1 deletion
+349 -1
README.md
+ 349
− 1
View file @ 157d169a
# Proj121-金角银角与钝角
Proj121-page-table-using-hashtable
使用哈希页表实现虚拟机的stage-2页表
\ No newline at end of file
使用哈希页表实现虚拟机的stage-2页表
## 一、内核设计
## 二、实验过程
### 第一题
### 第二题
### 第三题
#### 哈希页表的建立
> 在RISC-V Hypervisor的stage-2缺页流程中，实现哈希页表建立过程。
##### 建立页表的预置要求
修改两个工具函数：
* `my_set_pte`修改自`stage2_set_pte`，目的是为了将value值填入`pte`表项中。
```c
void my_set_pte(pte_t * ptep, unsigned long value){
//将ptep的前7个bytes置为value的值
unsigned char* charValue = (unsigned char*)&value;
unsigned char* charPte = (unsigned char*)ptep;
int i=0;
for(i=0;i<7;i++){
charPte[i] = charValue[i];
}
}
```
* `my_pte_val`修改自`pte_val`，目的是为了获取`pte`表项的值。
```c
unsigned long my_pte_val(pte_t pte){
//取出pte前7个bytes的值
unsigned long tmp = 0;
unsigned char* charPte = (unsigned char*)&pte;
unsigned char* charValue = (unsigned char*)&tmp;
int i=0;
for(i=0;i<7;i++){
charValue[i] = charPte[i];
}
return tmp;
}
```
##### 实现缺页填表操作
实现缺页填写页表操作：实现`hash`函数对于`addr`的映射以及开放寻址。
```c
static int stage2_set_pte(struct kvm *kvm, u32 level,
struct kvm_mmu_memory_cache *pcache, gpa_t addr,
const pte_t *new_pte)
{
//获取PGD的起始地址
unsigned char *tmpPgd = (unsigned char *)(kvm->arch.pgd);
//获取哈希的idx和组号
unsigned long blockNum = (unsigned long)addr >> 17;
unsigned long hashIdx = blockNum & 0xfff; //取后12位作为idx
unsigned long blockOff = (unsigned long)addr << 47 >> 61;
//获取该行的tag，需要进行比对是否一致
unsigned char *curLine = tmpPgd + (hashIdx * 64);
//前8位(再去掉最开始两个)为tag位
unsigned long tag, valid;
unsigned long i = 0, j = 0;
//startTable
unsigned char *startTable;
//PTE开始的地方
unsigned char *startPTE = (unsigned char *)&(new_pte->pte);
while (1) {
tag = *((unsigned long *)curLine);
startTable = curLine + blockOff * 7 + 8;
valid = tag >> 56;
if (valid) {
if ((tag & 0x0000ffffffffffff) == blockNum) {
//write PTE
my_set_pte((pte_t *)startTable,startPTE);
break;
} else {
//跳到下一行
curLine += 64;
}
} else {
//write PTE
my_set_pte((pte_t *)startTable,startPTE);
//write TAG
*((unsigned long *)curLine) =
0x0100000000000000 | blockNum;
break;
}
}
return 0;
}
```
##### 实现建表相关函数
* `stage2_op_pte`是负责初始化页表项的函数，需要将其修改为支持哈希页表版本。
```c
static void stage2_op_pte(struct kvm *kvm, gpa_t addr, pte_t *ptep,
u32 ptep_level, enum stage2_op op)
{
int i, ret;
pte_t *next_ptep;
u32 next_ptep_level;
unsigned long next_page_size, page_size;
page_size = PAGE_SIZE;
BUG_ON(addr & (page_size - 1));
//判断这个pte为0
if (!my_pte_val(*ptep))
return;
if (op == STAGE2_OP_CLEAR)
my_set_pte(ptep, 0);//定义函数:
else if (op == STAGE2_OP_WP)
my_set_pte(ptep, my_pte_val(*ptep) & ~_PAGE_WRITE);
}
```
* `stage2_get_leaf_entry`是根据获取`addr`对应页表项的函数，在多级页表中该函数会迭代查找，直到找到叶节点（存储最终的物理地址）。
```c
static bool stage2_get_leaf_entry(struct kvm *kvm, gpa_t addr, pte_t **ptepp,
u32 *ptep_level)
{
//addr ---hash---> 找到对应的位置
//获取PGD的起始地址
unsigned char *tmpPgd = (unsigned char *)(kvm->arch.pgd);
//获取哈希的idx和组号
unsigned long blockNum = (unsigned long)addr >> 17;
unsigned long hashIdx = blockNum & 0xff; //取后8位作为idx
unsigned long blockOff = (unsigned long)addr << 47 >> 61;
//获取该行的tag，需要进行比对是否一致
unsigned char *curLine = tmpPgd + (hashIdx * 64);
//前8位(再去掉最开始两个)为tag位
unsigned long tag, valid;
unsigned long i = 0, j = 0;
//startTable
unsigned char *startTable;
while (1) {
tag = *((unsigned long *)curLine);
startTable = curLine + blockOff * 7 + 8;
valid = tag >> 56;
if (valid) {
if ((tag & 0x0000ffffffffffff) == blockNum) {
//找到正确的位置
*ptep_level = 1;
*ptepp = (pte_t *)startTable;
break;
} else {
//跳到下一行
curLine += 64;
}
} else {
return false;
}
}
return true;
}
```
#### 哈希页表的查询
# 三、测试部分
我们在`stream`测试集上进行了测试，对比了在Host上的使用16kb页表大小与使用原有4kb页表大小的性能，以及在启用`kvm`使能后，在Guest上的使用hash页表与使用原有三级页表映射的性能。
## host-16k-hash
```bash
/apps # ./stream_c.exe
-------------------------------------------------------------
STREAM version $Revision: 5.10 $
-------------------------------------------------------------
This system uses 8 bytes per array element.
-------------------------------------------------------------
Array size = 10000000 (elements), Offset = 0 (elements)
Memory per array = 76.3 MiB (= 0.1 GiB).
Total memory required = 228.9 MiB (= 0.2 GiB).
Each kernel will be executed 10 times.
The *best* time for each kernel (excluding the first iteration)
will be used to compute the reported bandwidth.
-------------------------------------------------------------
Number of Threads requested = 1
Number of Threads counted = 1
-------------------------------------------------------------
Your clock granularity/precision appears to be 2 microseconds.
Each test below will take on the order of 187141 microseconds.
(= 93570 clock ticks)
Increase the size of the arrays if this shows that
you are not getting at least 20 clock ticks per test.
-------------------------------------------------------------
WARNING -- The above is only a rough guideline.
For best results, please be sure you know the
precision of your system timer.
-------------------------------------------------------------
Function Best Rate MB/s Avg time Min time Max time
Copy: 1833.0 0.087840 0.087290 0.088834
Scale: 860.3 0.192468 0.185988 0.226276
Add: 1107.3 0.217874 0.216739 0.225400
Triad: 939.3 0.258147 0.255508 0.275860
-------------------------------------------------------------[ 22.601770] random: fast init done
Solution Validates: avg error less than 1.000000e-13 on all three arrays
-------------------------------------------------------------
```
## guest-16k-hash
```bash
-------------------------------------------------------------
STREAM version $Revision: 5.10 $
-------------------------------------------------------------
This system uses 8 bytes per array element.
-------------------------------------------------------------
Array size = 10000000 (elements), Offset = 0 (elements)
Memory per array = 76.3 MiB (= 0.1 GiB).
Total memory required = 228.9 MiB (= 0.2 GiB).
Each kernel will be executed 10 times.
The *best* time for each kernel (excluding the first iteration)
will be used to compute the reported bandwidth.
-------------------------------------------------------------
Number of Threads requested = 1
Number of Threads counted = 1
-------------------------------------------------------------
Your clock granularity/precision appears to be 2 microseconds.
Each test below will take on the order of 262511 microseconds.
(= 131255 clock ticks)
Increase the size of the arrays if this shows that
you are not getting at least 20 clock ticks per test.
-------------------------------------------------------------
WARNING -- The above is only a rough guideline.
For best results, please be sure you know the
precision of your system timer.
-------------------------------------------------------------
Function Best Rate MB/s Avg time Min time Max time
Copy: 1364.9 0.121861 0.117227 0.146466
Scale: 645.2 0.250996 0.247974 0.261212
Add: 828.8 0.294635 0.289580 0.311800
Triad: 699.8 0.359421 0.342947 0.478321
-------------------------------------------------------------
Solution Validates: avg error less than 1.000000e-13 on all three arrays
-------------------------------------------------------------
```
## host-4k
```bash
-------------------------------------------------------------
STREAM version $Revision: 5.10 $
-------------------------------------------------------------
This system uses 8 bytes per array element.
-------------------------------------------------------------
Array size = 10000000 (elements), Offset = 0 (elements)
Memory per array = 76.3 MiB (= 0.1 GiB).
Total memory required = 228.9 MiB (= 0.2 GiB).
Each kernel will be executed 10 times.
The *best* time for each kernel (excluding the first iteration)
will be used to compute the reported bandwidth.
-------------------------------------------------------------
Number of Threads requested = 1
Number of Threads counted = 1
-------------------------------------------------------------
Your clock granularity/precision appears to be 2 microseconds.
Each test below will take on the order of 210880 microseconds.
(= 105440 clock ticks)
Increase the size of the arrays if this shows that
you are not getting at least 20 clock ticks per test.
-------------------------------------------------------------
WARNING -- The above is only a rough guideline.
For best results, please be sure you know the
precision of your system timer.
-------------------------------------------------------------
Function Best Rate MB/s Avg time Min time Max time
Copy: 1326.4 0.121248 0.120624 0.122517
Scale: 703.3 0.231843 0.227493 0.246709
Add: 854.4 0.287438 0.280907 0.333396
Triad: 770.8 0.313557 0.311359 0.324700
-------------------------------------------------------------
Solution Validates: avg error less than 1.000000e-13 on all three arrays
-------------------------------------------------------------
```
## guest-4k
```bash
-------------------------------------------------------------
STREAM version $Revision: 5.10 $
-------------------------------------------------------------
This system uses 8 bytes per array element.
-------------------------------------------------------------
Array size = 10000000 (elements), Offset = 0 (elements)
Memory per array = 76.3 MiB (= 0.1 GiB).
Total memory required = 228.9 MiB (= 0.2 GiB).
Each kernel will be executed 10 times.
The *best* time for each kernel (excluding the first iteration)
will be used to compute the reported bandwidth.
-------------------------------------------------------------
Number of Threads requested = 1
Number of Threads counted = 1
-------------------------------------------------------------
Your clock granularity/precision appears to be 2 microseconds.
Each test below will take on the order of 313721 microseconds.
(= 156860 clock ticks)
Increase the size of the arrays if this shows that
you are not getting at least 20 clock ticks per test.
-------------------------------------------------------------
WARNING -- The above is only a rough guideline.
For best results, please be sure you know the
precision of your system timer.
-------------------------------------------------------------
Function Best Rate MB/s Avg time Min time Max time
Copy: 640.3 0.251023 0.249885 0.253838
Scale: 419.9 0.383085 0.381044 0.388234
Add: 476.1 0.509056 0.504084 0.538106
Triad: 440.2 0.550963 0.545153 0.586988
-------------------------------------------------------------
Solution Validates: avg error less than 1.000000e-13 on all three arrays
-------------------------------------------------------------
```
<img src="https://s2.loli.net/2022/05/12/bdh9Ikptz1i86gu.png" alt="Guest" style="zoom:67%;" />
<img src="https://s2.loli.net/2022/05/12/MYghpAPsboaS1ew.png" alt="Host" style="zoom:67%;" />
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment