深入理解計(jì)算機(jī)系統(tǒng)第二版資料-家庭作業(yè)答案(參考版)
2025-06-23 00:46本頁面
【正文】 dst數(shù)組 列0列1列2列3行0mhmh行1mmh。src[0] src[2] 對(duì)應(yīng)組0；src[1] src[3] 對(duì)于組1。1211109876543210CTCTCTCTCTCTCTCTCICICICOCO12111098765432100011100011000B.參數(shù)值塊偏移CO0x00索引CI0x6標(biāo)記CT0x38命中？Yes返回值0xFA12111098765432101011011101100B.參數(shù)值塊偏移CO0x00索引CI0x3標(biāo)記CT0xB7命中？No返回值0x13140x13170x17940x1797對(duì)于寫分配的高速緩存，每次寫不命中時(shí)，需要讀取數(shù)據(jù)到高速緩存中?！⌒?行2組00x01200x0123組10x08A40x08A70x07040x0707組2組30x064C0x064F組40x18F00x18F30x00B00x00B3組50x0E340x0E37組60x12380x123B組70x1BDC0x1BDFb = 2, s = 2, t = 12 b s = 8。Address_start = (tag5) | (set2)。T = T_seek + *T_maxrotate + 6*T_maxrotate = 那么3MB文件就有3000個(gè)邏輯塊（扇區(qū)），需要讀6個(gè)磁道。 msT_transfer = T_maxrotate/500 =T_seek =2*pi*x*r*(1x)*K，其中pi，r和K都是常數(shù)，那么只剩下x*(1x)。假設(shè)半徑為r的磁盤總的磁道是K，那么除掉內(nèi)部的x*r（磁道數(shù)為x*K），剩下的磁道數(shù)為 (1x)*K。所以提高B會(huì)使得性能更優(yōu)。將B提到3倍，也就是B需要10T，那么總時(shí)間為80T。= v。a[i]。v = v{i++)in。for(。}v2。 v = v p[i+1]p[i]+v2 = a[i+1]。v1 = a[i]。v = 0。i。longn){longfloatpsum(float}關(guān)鍵路徑就是一個(gè)浮點(diǎn)數(shù)乘法，因此CPE是浮點(diǎn)乘法延遲的1/4，然而每次計(jì)算都需要load 4個(gè)值。return}powx4s{++i)= degree。*= x4。powx4。v1s = s*+a[i+3]*x。v2 = a[i+2]a[i+1]*x。v1 = a[i]+=i0。limit = degree3。longdoubledoublepowx4 =s =0。doubleintintdoublepoly_optimized(doubles。n。(unsigned*schar++{//剩余的nlchar。(unsignedn*lchar++{= K(schar。(unsignedn。(unsigned*schar++{n)amp。while((unsigned)schar%K(K1)來求schar%K(i3)。0xff)+=for(i =i =0。long*lchar。unsigned(unsignedcharlong)。K =unsignedsize_t n){int*optimized_memset(void*dest = sum。*+x3)。x2(x1sum = sum*int*int*int{(i =2。limit = length0。data_t sum =*vdata =get_vec_start(u)。data_tlength =i。long*dest){vec_ptr v,voidB. 關(guān)鍵路徑上仍然有N個(gè)浮點(diǎn)加法，所以循環(huán)展開并沒有改變A. load執(zhí)行單元的吞吐量B. IA32可用寄存器實(shí)際只有6個(gè)，而三路展開需要i, limit, udata, vdata，以及存儲(chǔ)udata[i], vdata[i]的寄存器，所以肯定有些循環(huán)變量會(huì)溢出到寄存器，這會(huì)影響效率。*dest = sum。*+vdata[i+2]。udata[i+2]sum = sum*+vdata[i]。udata[i]sum = sum{(i =2。limit = length0。data_t sum =*vdata =get_vec_start(u)。data_tlength =i。long*dest){vec_ptr v,add+020load2+13load1add+14load2+25load1add+26mul(load延遲4)load2...7...80mul9add整數(shù)mul延遲為3010added整數(shù)加法延遲為1mul11add12addedvoid以整數(shù)為例：相同底色表示這些指令在一個(gè)循環(huán)內(nèi)執(zhí)行，以及同一個(gè)循環(huán)內(nèi)的初始值：，？浮點(diǎn)數(shù)的話。整個(gè)程序的限制因素為最后的浮點(diǎn)數(shù)加法的延遲。C. 兩個(gè)load操作的吞吐量界限。深入理解計(jì)算機(jī)系統(tǒng)(第二版) 家庭作業(yè) 第五章 A.關(guān)鍵路徑是%xmm1更新路徑上的加法。暫時(shí)沒有想到好的辦法。版本1，在預(yù)測正確的情況下執(zhí)行7條指令，預(yù)測錯(cuò)誤時(shí)執(zhí)行9條指令并插入一個(gè)bubble。 1 : E_valA。amp。 D_icode == IPUSHL)。!(E_icode == IMRMOVL amp。amp。amp。 感覺蠻麻煩的，不想寫啊。ILEAVE Other instructions don’t need address]。icode in { IPOPL, IRET } : valA。}。bool set_cc = icode in { IOPL,icode in { IRRMOVL, IIRMOVL}:0。ILEAVE, IADDL Other instructions don’t need ALU]。icode in { IRET, IPOPL,} : valC。icode in { IIRMOVL, IRMMOVL, IMRMOVL,執(zhí)行int aluA = [icode in { IRRMOVL, IOPL1 : RNONE。 icode in { IMRMOVL, IPOPL}:rA。int dstM = [1 : RNONE。icode in { IPUSHL, IPOPL, ICALL, IRET,IADDL: rB。amp。int dstE = [icode in { IRRMOVL}1 : RNONE。icode in { IPUSHL, IPOPL, ICALL, IRET } : RESP。IADDL Don’t need register]。IRET } : RESP。icode in {icode in { IRRMOVL, IRMMOVL, IOPL, IPUSHL譯碼和寫回int srcA = [IADDL}。IIRMOVL, IRMMOVL, IMRMOVL,icode in { IRRMOVL, IOPL, IPUSHL, IPOPL,取指bool need_regids =R[rB] = valE SetCC訪存階段 valE = valB + valC valP = PC + 6譯碼階段 valC = M4[PC+2] rA:rB = M1[PC+1]icode:ifun = M1[PC] = C:0 R[%esp] = valE訪存階段 valE = valB + 4 valP = PC + 1譯碼階段 InnerLoop我們可以明確的是，這條指令完成的任務(wù)為，ebp M4[cur_ebp]esp cur_ebp + 4取指階段 icode:ifun = D:0%eax, %ecxmovl $1, %eaxsubl%edx, 4(%ecx)movl edx is maxmovl (%ecx), %eax pare *p and *(p+1)cmovl4(%ecx), %ebxsubl(%ecx), %edxInnerLoop:movl%ebp, %esppopl%ebxpoplOuterLoopbubbleEnd:popl%esimovl $1, %eaxsubl%eax, %ecxmovl $1, %eaxsubl%eax, %esijneswap, so ebx is greatermovl 4(%ecx), %eaxNoSwap:movlNoSwapmovl4(%ecx), %ebxsubl%esi to save one regInnerLoop:movl%edx, %ecxbubbleEndcount==0movl $1, %eaxsubl%esi, %esije12(%ebp), %esi 8(%ebp), %edx %ebxpushl%esp, %ebppushlcount = 8movl data, %ecxpushl %ecx push datacall bubblehaltbubble:pushl)data:（從地地址往高地址）$5, $2, $7, $4, $3, $6, $1, $8movl $8, %ecxq++。*q = t。*p =t =*q*p1。q = data*q。inti,return。if(count ==count){*data,void如果REG是esp，那么最后得到esp是棧頂值減4之后的值。如果REG是esp，那么代碼是先減去了esp，然后將減了4以后的REG移入了esp。}函數(shù)的目的是找到樹的所有節(jié)點(diǎn)的值中最小的一個(gè)。return}?//Line16 cmovle: if(r12rax) rax=r12。lv : rvrvv = lvrv =lv ={!= NULL)lv。v = MAX_LONG,}當(dāng)然，如果要用三目條件表達(dá)式的話：longreturn}min(v,//Line16 cmovle: if(r12rax) rax=r12。traverse(tpright))。v ={!= NULL)v = MAX_LONG。longret。tp = tpleft。while(tpret =long需要在最開始輸入的時(shí)候按，即按了回車之后按。}兩種方法對(duì)于EOF好像沒效果，就是輸入一定字符后不按回車直接按EOF，沒能正確輸出。break。putchar(s[i])。s[i]。BufferSize,i。s[Buffe
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