译注:本章基本未做翻译

10.6 TLB Testing TLB测试

The 80386 provides a mechanism for testing the Translation Lookaside Buffer (TLB), the cache used for translating linear addresses to physical addresses. Although failure of the TLB hardware is extremely unlikely, users may wish to include TLB confidence tests among other power-up confidence tests for the 80386.

80386提供一个测试转换旁路缓冲(TLB)的机制,这个缓冲用来将线性地址转换为物理地址。尽管TLB硬件失效是极端行为,用户还是希望能对80386进行包括TLB信任测试的其他加电信任测试。

Note

This TLB testing mechanism is unique to the 80386 and may not be continued in the same way in future processors. Sortware that uses this mechanism may be incompatible with future processors.

When testing the TLB it is recommended that paging be turned off (PG=0 in CR0) to avoid interference with the test data being written to the TLB.

10.6.1 Structure of the TLB

The TLB is a four-way set-associative memory. Figure 10-3 illustrates the structure of the TLB. There are four sets of eight entries each. Each entry consists of a tag and data. Tags are 24-bits wide. They contain the high-order 20 bits of the linear address, the valid bit, and three attribute bits. The data portion of each entry contains the high-order 20 bits of the physical address.

10.6.2 Test Registers

Two test registers, shown in Figure 10-4, are provided for the purpose of testing. TR6 is the test command register, and TR7 is the test data register. These registers are accessed by variants of the MOV instruction. A test register may be either the source operand or destination operand. The MOV instructions are defined in both real-address mode and protected mode. The test registers are privileged resources; in protected mode, the MOV instructions that access them can only be executed at privilege level 0. An attempt to read or write the test registers when executing at any other privilege level causes a general protection exception.

The test command register (TR6) contains a command and an address tag to use in performing the command:

C

This is the command bit. There are two TLB testing commands: write entries into the TLB, and perform TLB lookups. To cause an immediate write into the TLB entry, move a doubleword into TR6 that contains a 0 in this bit. To cause an immediate TLB lookup, move a doubleword into TR6 that contains a 1 in this bit.

Linear Address

On a TLB write, a TLB entry is allocated to this linear address; the rest of that TLB entry is set per the value of TR7 and the value just written into TR6. On a TLB lookup, the TLB is interrogated per this value; if one and only one TLB entry matches, the rest of the fields of TR6 and TR7 are set from the matching TLB entry.

V

The valid bit for this TLB entry. The TLB uses the valid bit to identify entries that contain valid data. Entries of the TLB that have not been assigned values have zero in the valid bit. All valid bits can be cleared by writing to CR3.

D, D#

The dirty bit (and its complement) for/from the TLB entry.

U, U#

The U/S bit (and its complement) for/from the TLB entry.

W, W#

The R/W bit (and its complement) for/from the TLB entry.

The meaning of these pairs of bits is given by Table 10-1, where X represents D, U, or W.

The test data register (TR7) holds data read from or data to be written to the TLB.

Physical Address

This is the data field of the TLB. On a write to the TLB, the TLB entry allocated to the linear address in TR6 is set to this value. On a TLB lookup, if HT is set, the data field (physical address) from the TLB is read out to this field. If HT is not set, this field is undefined.

HT

For a TLB lookup, the HT bit indicates whether the lookup was a hit (HT := 1) or a miss (HT := 0). For a TLB write, HT must be set to 1.

REP

For a TLB write, selects which of four associative blocks of the TLB is to be written. For a TLB read, if HT is set, REP reports in which of the four associative blocks the tag was found; if HT is not set, REP is undefined.

Table 10-1. Meaning of D, U, and W Bit Pairs

X     X#      Effect during        Value of bit X

TLB Lookup           after TLB Write

0     0       (undefined)          (undefined)

0     1       Match if X=0         Bit X becomes 0

1     0       Match if X=1         Bit X becomes 1

1     1       (undefined)          (undefined)

10.6.3 Test Operations

To write a TLB entry:

  1. Move a doubleword to TR7 that contains the desired physical address, HT, and REP values. HT must contain 1. REP must point to the associative block in which to place the entry.
  2. Move a doubleword to TR6 that contains the appropriate linear address, and values for V, D, U, and W. Be sure C=0 for "write" command.

Be careful not to write duplicate tags; the results of doing so are undefined. To look up (read) a TLB entry:

  1. Move a doubleword to TR6 that contains the appropriate linear address and attributes. Be sure C=1 for "lookup" command.
  2. Store TR7. If the HT bit in TR7 indicates a hit, then the other values reveal the TLB contents. If HT indicates a miss, then the other values in TR7 are indeterminate.

For the purposes of testing, the V bit functions as another bit of addresss. The V bit for a lookup request should usually be set, so that uninitialized tags do not match. Lookups with V=0 are unpredictable if any tags are uninitialized.

【译】x86程序员手册41-10.6 TLB(快表)测试的更多相关文章

  1. 【译】x86程序员手册01

    Intel 80386 Reference Programmer's Manual 80386程序员参考手册 Chapter 1 -- Introduction to the 80386 第1章 - ...

  2. 【译】x86程序员手册03 - 2.1内存组织和分段

    2.1 Memory Organization and Segmentation 内存组织和分段 The physical memory of an 80386 system is organized ...

  3. 【译】x86程序员手册10 - 第4章系统架构

    1.1.2 Part II -- Systems Programming 系统编程 This part presents those aspects of the architecture that ...

  4. 【译】x86程序员手册38-10.2实在址模式下的软件初始化

    10.2 Software Initialization for Real-Address Mode   实地址模式的软件初始化 In real-address mode a few structur ...

  5. 【译】x86程序员手册11- 4.1系统寄存器

    4.1 Systems Registers 系统寄存器 The registers designed for use by systems programmers fall into these cl ...

  6. 【译】x86程序员手册00 - 翻译起因

    从上一次学习MIT的操作系统课程又过去了一年.上次学习并没有坚持下去.想来虽有种种原因,其还在自身无法坚持罢了.故此次再鼓起勇气重新学习,发现课程都已由2014改版为2016了.但大部分内容并没有改变 ...

  7. 【译】x86程序员手册40-10.5初始化的例子

    10.5 Initialization Example初始化的例子 译注:本来想把这个例子全部注释完,但由于对intel汇编实不熟悉,有太多的伪指令,本人也是免强看懂,所以就不再做翻译了. $TITL ...

  8. 【译】x86程序员手册39-10.3切换到保护模式

    10.3 Switching to Protected Mode  切换到保护模式 Setting the PE bit of the MSW in CR0 causes the 80386 to b ...

  9. 【译】x86程序员手册37-第10章 初始化

    Chapter 10 Initialization 第10章 初始化 After a signal on the RESET pin, certain registers of the 80386 a ...

随机推荐

  1. java Http post请求发送json字符串

    最近差点被业务逻辑搞懵逼,果然要先花时间思考,确定好流程再执行.目前最好用的jar包还是org.apache.http. public class HttpClientHelper { private ...

  2. Lightoj 1018 - Brush (IV)

    1018 - Brush (IV)    PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Muba ...

  3. YTU 2573: 连续奇数和

    2573: 连续奇数和 时间限制: 1 Sec  内存限制: 128 MB 提交: 63  解决: 37 题目描述 小明看到一本书上写着:任何数字的立方都可以表示为连续奇数的和. 比如: 2^3 = ...

  4. Resolve the error: an error occurred during local report processing

    Issue: an error occurred during local report processing.the definition of the report'Main Report'is ...

  5. 【SCOI 2005】 最大子矩阵

    [题目链接] 点击打开链接 [算法] 动态规划 我们发现,M只有两种取值,1和2,那么,只需分类讨论即可 当M = 1时,其实这个问题就成了就最大连续子段和的问题,只不过要选K段而已 用f[i][j] ...

  6. Educational Codeforces Round 22 补题 CF 813 A-F

    A The Contest 直接粗暴贪心 略过 #include<bits/stdc++.h> using namespace std; int main() {//freopen(&qu ...

  7. 思维导图js

    http://baike.baidu.com/view/1469162.htm?fromenter=Mind+Maps

  8. 小程序-demo:小程序示例-page/component

    ylbtech-小程序-demo:小程序示例-page/component 以下将展示小程序官方组件能力,组件样式仅供参考,开发者可根据自身需求自定义组件样式,具体属性参数详见小程序开发文档. 1. ...

  9. HDU 2340 Obfuscation (暴力)

    题意:给定一篇文章,将每个单词的首尾字母不变,中间顺序打乱,然后将单词之间的空格去掉,得到一个序列,给出一个这样的序列,给你一个字典,将原文翻译出来. 析:在比赛的时候读错题了,忘记首尾字母不变了,一 ...

  10. Spring @requestBody

    页面提交请求参数有两种,一种是form格式,一种是json格式 jQuery的$.post方法虽然也可以传递json格式数据,但实际上是用的form格式提交,jquery会帮你把json转成form格 ...