計算機系統(tǒng)基礎(chǔ)

計算機系統(tǒng)基礎(chǔ)Computer

System

Fundamentals想聽什么？你來定義！sh@歡迎入群，一起

，共同定義課程！Class

1計算機系統(tǒng)概述@sh

注:本節(jié)內(nèi)容大量參考《Computer

Systems:A

Programmer’s

Perspec<ve》一書及課件為什么要上這門課?課程總體介紹信息的表示AgendaSource:/571/*

Kernel

memory

region

holding

user-accessible

data

*/#define

KSIZE

1024char

kbuf[KSIZE];size_t

memcpy(void*,

void*,

size_t);/*

Copy

at

most

maxlen

bytes

from

kernel

region

to

user

buffer

*/int

copy_from_kernel(void

*user_dest,

int

maxlen){/*

Byte

count

len

is

minimum

of

buffer

size

and

maxlen

*/int

len

=

KSIZE

<

maxlen

?

KSIZE

:

maxlen;memcpy(user_dest,

kbuf,

len);return

len;}這段代碼有什么問題?/*

Kernel

memory

region

holding

user-accessible

data

*/#define

KSIZE

1024char

kbuf[KSIZE];size_t

memcpy(void*,

void*,

size_t);/*

Copy

at

most

maxlen

bytes

from

kernel

region

to

user

buffer

*/int

copy_from_kernel(void

*user_dest,

int

maxlen)

{/*

Byte

count

len

is

minimum

of

buffer

size

and

maxlen

*/int

len

=

KSIZE

<

maxlen

?

KSIZE

:

maxlen;memcpy(user_dest,

kbuf,

len);return

len;}如果這么用呢?#define

MSIZE

528void

getstuff()

{char

mybuf[MSIZE];copy_from_kernel(mybuf,

-MSIZE);.

.

.}了解你的系統(tǒng)

–1Q1:

Is

x2

≥

0?–

Float’s:

Yes!–

Int’s:40000

*

40000

?

160000000050000

*

50000

?

??Q2:

Is

(x

+

y)

+

z

= x

+

(y

+z)?Unsigned

&

Signed

Int’s:

Yes!Float’s:(1e20

+

-‐1e20)

+

3.14

?

3.141e20

+

(-‐1e20

+3.14)

?

??int不是整數(shù)，?oat不是實數(shù)double

fun(int

i){volatile

double

d[1]

=

{3.14};volatile

long

int

a[2];a[i]

=

1073741824;

/*

Possibly

out

of

bounds

*/return

d[0];}假設(shè)In

x86架構(gòu)fun(0)fun(1)fun(2)fun(3)fun(4)?????？？？？？什么返回值?double

fun(int

i){volatile

double

d[1]

=

{3.14};volatile

long

int

a[2];a[i]

=

1073741824;

/*

Possibly

out

of

bounds

*/return

d[0];}假設(shè)In

x86架構(gòu)fun(0)fun(1)fun(2)fun(3)fun(4)?

3.14?

3.14?

3.1399998664856?

2.00000061035156?

3.14,

then

segmentation

fault返回值double

fun(int

i){volatile

double

d[1]

=

{3.14};volatile

long

int

a[2];a[i]

=

1073741824;

/*

Possibly

out

of

bounds

*/return

d[0];}fun(0)fun(1)fun(2)fun(3)fun(4)?

3.14?

3.14?

3.1399998664856?

2.00000061035156?

3.14,

then

segmentation

faultLoca<on

accessed

byfun(i)Explana<on:Saved

Stated7

...

d4d3

...

d0a[1]a[0]43210原因了解你的系統(tǒng)

–2內(nèi)存需要被管理和分配堆vs.棧在C/C++中內(nèi)存是不受保護(hù)的數(shù)組越界無效指針內(nèi)存錯誤可能導(dǎo)致非常詭異的問題癥狀跟系統(tǒng)和編譯器的實現(xiàn)相關(guān)可能延遲很久才出現(xiàn)借助內(nèi)存

檢查工具

(e.g.

valgrind)哪個比較快?void

copyji(int

src[2048][2048],intdst[2048][2048]){int

i,j;for

(j

=

0;

j

<

2048;

j++)for

(i

=

0;

i

<

2048;

i++)dst[i][j]

=

src[i][j];}void

copyij(int

src[2048][2048],intdst[2048][2048]){int

i,j;for

(i

=

0;

i

<

2048;

i++)for

(j

=

0;

j

<

2048;

j++)dst[i][j]

=

src[i][j];}Pen<um

4上性能相差21倍！數(shù)組:從概念到現(xiàn)實A[2047][2047]A[2047][2046]…A[2047][1]A[2047][0]…A[1][2047]A[1][2046]…A[1][1]A[1][0]A[0][2047]A[0][2046]…A[0][1]A[0][0]x

+

4xx

+

8192x

+

8188x

+

8184數(shù)組的內(nèi)存布局MainmemoryI/ObridgeBusinterfaceALURegister?leCPU

chipSystem

busMemory

busCachememories典型的系統(tǒng)結(jié)構(gòu)L1

CacheL2

CacheMemory110100SRAMSRAMDRAM位置延遲(cycles)介質(zhì)Registers(SRAM)Main

memory(DRAM)Larger,slower,cheaperper

byteRemote

secondarystorage(tapes,

distributed

?le

systems,

Web

servers)Local

disks

hold

?lesretrieved

from

disks

on

remote

network

serversMain

memory

holds

diskblocksretrieved

from

local

disksL2

cache(SRAM)L1

cache

holds

cache

lines

retrieved

from

L2

cacheCPU

registers

hold

words

retrievedfrom

L1

cacheL2

cache

holds

cache

linesretrieved

from

main

memoryL0:L1:

L1

cacheL2:L3:L4:Local

secondary

storage(local

disks)L5:Smaller,faster,costlierper

byte典型的Memory

Hierarchy8M1M128K16K2K01000200030006000500040007000s1s3s5s7s9s11s13s15s3264MSize

(bytes)Readthroughput

(MB/s)Stride

(x8

bytes)L1L2MemL3copyijcopyjiIn Core

i72.67

GHz32

KB

L1

d-cache256

KB

L2

cache8

MB

L3

cacheThe

Memory

Mountain了解你的系統(tǒng)

–3內(nèi)存的性能不是均勻的cache和虛擬內(nèi)存特性對性能有極大的影響使程序適應(yīng)于內(nèi)存的特性即使算法復(fù)雜度相同，程序性能可能仍有幾十倍的差異Multiple

threads:

4xVector

instructions:

4xMemory

hierarchy

and

other

optimizations:

20xLinux時鐘中斷周期:

10ms，#de?ne

HZ

100采用x86硬件寄存器Time

Stamp

Counter記錄CPU執(zhí)行的cycle數(shù)指令:rdtsc如何精確計時

–1uint64

t1,

t2;t1

=get_counter();func();t2

=get_counter();printf(“func

required

%u64

clock

cycles,

%f

seconds\n",t2-t1,

(double)(t2-t1)/FREQUENCY);如何精確計時

–2/*

Set

*hi

and

*lo

to

the

high

and

low

order

bits* of

the

cycle

counter.*/void

access_counter(uint32

*hi,

uint32

*lo){asm("rdtsc;

movl

%%edx,%0;

movl

%%eax,%1":

"=r"

(*hi),

"=r"

(*lo)::

"%edx",

"%eax");}uint64

get_counter(){uint64

t;access_counter(((uint32*)&t)+1,

(uint32*)&t);return

t;}如何高效拷貝文件

–1/*

Note:

to

simplify

the

implementation,*

no

failure

handling

is

provided.*/void

copyfile(FILE

*dst_fp,

FILE

*src_fp,int

size){char

buf[4096];while

(size

>

4096)

{fread(buf,

4096,

1,

src_fp);fwrite(buf,

4096,

1,

dst_fp);size

-=

4096;}if

(size

>

0)

{fread(buf,

size,

1,

src_fp);fwrite(buf,

size,

1,

dst_fp);}}diskmemkernelspaceuserspacesrcdstbufFSBu?erFILE

stream

Bu?er如何高效拷貝文件

–2void

mmapcopy(int

dst_fd,

int

src_fd,

int

size){char

*bufp;bufp

=

mmap(NULL,

size,

PROT_READ,

MAP_PRIVATE,

src_fd,

0);write(dst_fd,

bufp,

size);}diskmemkernelspaceuserspacesrcdstbufpFSBu?ervoid

*mmap(void

*start,

int

len,int

prot,

int

flags,

int

fd,

int

offset)；len

bytesstart(or

addresschosen

bykernel)Process

virtual

memoryDisk

?le

speci?ed

by?le

descriptor

fdlenbytesoffset(bytes)00mmap如何高效傳輸文件–1/*

Note:

to

simplify

the

implementation,*

no

failure

handling

is

provided.*/void

tranfile(int

sock,

int

fd,int

size){char

buf[4096];while

(size

>

4096)

{read(fd,

buf,

4096);write(sock,

buf,

4096);size

-=

4096;}if

(size

>

0)

{read(fd,

buf,

size);write(sock,

buf,

size);}}Network

StackdiskmemkernelspaceuserspacefdbufFSBu?ersock如何高效傳輸

文件

–

2ssize_dfile(int

sock,

int

fd,off_t

*offset,

size_t

count);Network

StackdiskmemkernelspaceuserspacefdFSBu?ersockkernel

2.1Network

StackdiskmemkernelspaceuserspacefdFSBu?ersockkernel2.4你需要知道的還有很多CPU的工作機理計算機的功能部件與特點OS的工作機理編譯器的實現(xiàn)軟硬件如何配合、各

層次如何配合通訊與網(wǎng)絡(luò)編程模型OS體系結(jié)構(gòu)為什么要上這門課?課程總體介紹信息的表示Agenda課程內(nèi)容涉及5大基礎(chǔ)領(lǐng)域Computer

ArchitectureOpera<ng

SystemsProgramming

ModelNetworkingDistributed

Systems目標(biāo)：理解你看到的系統(tǒng)從場景和實例出發(fā)介紹概念兼顧應(yīng)用

原理

(程序員視角

+

系統(tǒng)工程師視角)體系化，強調(diào)系統(tǒng)全景內(nèi)容組織方式Applica<onDesign

&Implementa<onPrinciplesWhat:

50%How:

30%Why:

20%理解你的系統(tǒng)：1.

實例+概念2.應(yīng)用+原理3.

體系化為什么要上這門課?課程總體介紹信息的表示Agenda3.3V2.8V0.5V0.0V010二進(jìn)制的物理表示簡單?性強Byte

=

8

bits–2進(jìn)制:000000002

~

111111112–10進(jìn)制:010

~25510–16進(jìn)制:0016

~

FF16‘0’

~

‘9’,

‘A’

~

‘F’FA1D37B16

在C語言中的表示0xFA1D37B0xfa1d37b000000110001220010330011440100550101660110770111881000991001A101010B111011C121100D131101E141110F151111用bit組成byte邏輯上是一個大“字節(jié)數(shù)組”虛擬內(nèi)存地址空間進(jìn)程私有w連續(xù):0~2-‐1，但有空洞虛實動態(tài)分配:malloc多種形式:sta<c,stack,heap實際–

Memory

Hierarchy?

?

?內(nèi)存以byte組成RegistersL1

cache(SRAM)Main

memory(DRAM)Local

secondary

storage(localdisks)Remote

secondarystorage(tapes,

distributed

?le

systems,

Web

servers)L2

cache(SRAM)L0:L1:L2:L3:L4:L5:字長(word

size)操作數(shù)寬度，地址寬度一般機器采用32位字長(4bytes)地址空間4G高端機器采用64位字長(8bytes)地址空間1.8x1019x86-‐64使用48位地址:256TB多種操作數(shù)寬度1,

2,

4,

8

bytes計算機的字

(word)CData

TypeTypical

32-bitIn

IA32x86-64char111short222int444long448long

long888float444double888long

double810/1210/16pointer448數(shù)據(jù)類型的寬度printf(”%d\n”,

sizeof(long));一個word在內(nèi)存中如何以byte存放?在字節(jié)流中如何排列?交換數(shù)據(jù)時必須一致兩個傳統(tǒng)大尾端(Big

Endian):Sun,PPC

Mac,

InternetLSB

(Least

Signi?cant

Byte)在高地址小尾端(Livle

Endian):x86LSB在低地址字節(jié)序

(byte

ordering)假設(shè)int

x=0x01234567，&x=0x100大尾端–LSB在高地址(閱讀序)0x100

0x1010x102

0x10301234567小尾端–LSB在低地址(逆閱讀序)0x100

0x101

0x102

0x10367452301字節(jié)序的例子MSB

LSB起始地址，漸增4Bytes匯編二進(jìn)制機器碼的文本表示，語義上等價可由程序自動翻譯,objdump例子值:填滿32bits(padding):按byte分解:倒序:0x12ab0x000012ab00

00

12

abab

12

00

00地址8048365:8048366:機器碼5b81

c3ab120000匯編表示

popadd%ebx$0x12ab,%ebx804836c:83

bb2800000000cmpl$0x0,0x28(%ebx)?如何讀數(shù)看一段匯編代碼從byte到數(shù)據(jù)類型整數(shù)

int浮點數(shù)?oat字符串string指針pointer可把任意數(shù)據(jù)類型轉(zhuǎn)化為字節(jié)數(shù)組來觀測–用unsigned

char*指向其地址typedef

unsigned

char

*pointer;void

show_bytes(pointer

start,

int

len){int

i;for

(i

=

0;

i

<

len;

i++)printf(”%p\t0x%.2x\n",

start+i,

start[i]);printf("\n");}show_bytes例子int

a

=

15213;printf("int

a

=

15213;\n");show_bytes((pointer)&a,

sizeof(int));Result

(Linux):int

a

=

15213;0x11ffffcb8

0x6d0x11ffffcb9

0x3b0x11ffffcba

0x000x11ffffcbb

0x00Decimal:Binary:Hex:152130011

1011

0110

11013

B

6

D6D3B0000IA32,

x86-6400003B6DSunint

a

=

15213;93C4FFFFIA32,

x86-64FF