教程講稿與匯編

預(yù)備知識1K

=1M=1024K=1G=1024M=1T=1024G=1P=1024T=

1個二進(jìn)制位 （比特8個二進(jìn)制位：Byte（字節(jié)）1Byte=8bit Word（字）1Word=2Byte=數(shù)制基數(shù) 二進(jìn)20,八進(jìn)80,1,2,3,4,5,6,十進(jìn)0,1,2,3,4,5,6,7,8,十六進(jìn)0,1,2,3,4,5,6,7,8,A,B,C,D,E, 降冪法 05C3

3D25＋3＋31D2EHH—4205C376HH“與”運算 “或”運算 AB AB000000010011100101111111 AAAB0100010011101110例：X= Y=5555H，Z=XY=X=0000000011111111Y=0101010101010101Z=0101010110101010Z=機器字(machineword)地址的表示空間是對很多地址的表示空間是對很多內(nèi)存需求量大的應(yīng)用而言，非常有64位字地址的表示空間約是1.8X1019目前的x86-64機型實際支持48位寬的地址:256==地====

=

=== BigEndian:Sun,PPCMac,低位字節(jié)(Leastsignificantbyte,LSB)占據(jù)高地LittleEndian:與LSB相Big

數(shù)值是 Little

001110110011101101103B6DintB=-longintC=IA32,x86-

x86-

IA32,x86- CDataTypicalCDataTypical32-x86-x86-111222444448long888444888long8char448Oranyother 判斷以下的推斷或者等式是否成立(不成立則給出范例x,y為32位帶符號整初始

x< ((x*2)<ux>=x&7== (x<<30)<ux>-intx=inty=bar();unsignedux=unsignedintx=inty=bar();unsignedux=unsigneduy=x*x>=x>0&&y> x+y>x>= -x<=x<= -x>=(x|-x)>>31==-ux>>3==xx>>3==x&(x-1)!= 無符號 帶符號數(shù)（補碼，Two’sB2U(X xi2

B2T(X xw12w1xishortshortintx=15213;shortinty=-x3By-C4符號位（sign對于補碼表示,MSB表示整數(shù)的符1for01for無符號

帶符號數(shù)（補碼

2w–

2w–1–負(fù)1=假設(shè)字長為FF7F-80-FF0000XX001122334455667789

xwx x常數(shù)如果有“U”作為后綴則是無符號數(shù)，如 如果無符號數(shù)與帶符號數(shù)混合使用，則帶符號數(shù)默認(rèn)轉(zhuǎn)換為無符號UU實例UU00-U-(unsigned)-UMax–1

TMax+

模運按位運建議：不能僅僅因為取值范圍是非負(fù)而使forfor(i=cnt-2;i>=0;i--a[i]+=#defineDELTAsizeof(int)inti;for(i=CNT;i-DELTA>=0;i-=..TrueSum:w+1

?????????u+??DiscardCarry:w UAddw(u?? UAddw(u, （u+v）mod uwuvuv2uv2TrueSum:w+1

?????????u+??DiscardCarry:w TAddw(u??與無符號數(shù)的一Signedvs.unsignedadditioninints,t,u,s=(int)((unsigned)u+(unsigned)t=u+s==0

True0

2w

TAdd0

11

–2w

v2w

uvTMinwuvTMaxwuu>>kgivesu/2k采用邏輯右00uBinary00uBinary/u/.u/2k000100xBx>>1Dx>>03x>>00Cunsignedunsignedudiv8(unsigned{return}CompiledArithmetic $3, $3,#Logicalshiftreturnx>>x>>kgivesx/2k采用算術(shù)右但是x<0時，舍入錯000

Binary110000/x00/x/.

RoundDown(x/y--C4y>>--E2y>>--FCy>>--FFWantx2k (需要向0舍入，而不是向下舍入Computeas(x+2k-1)/2kInC:(x+(1<<k)-1)>>BiasesdividendtowardCase1:No

110001010001110111Binary/u/2k001000111.11BiasingBiasinghasnoCase2:

k110110011001100

Incrementedby

Binary100/x/2k100/x/2k11.100

IncrementedbyCintintidiv8(int{return}CompiledArithmetic testltestl%eax,$3,$7,%eaxifx<0x+=#Arithmeticreturnx>> 判斷以下的推斷或者等式是否成立(不成立則給出示例x,y為32位帶符號整初始

x< ((x*2)<ux>=x&7== (x<<30)<ux>-intx=inty=bar();unsignedux=unsignedintx=inty=bar();unsi