C語(yǔ)言編程計(jì)算L1緩存

1、用C語(yǔ)言推測(cè)Cache相關(guān)參數(shù)2014-11-11 23:13 373 人閱讀 評(píng)論（0）收藏 舉報(bào)版權(quán)聲明：本文為博主原創(chuàng)文章，未經(jīng)博主允許不得轉(zhuǎn)載。目錄+1）塊大小思路：假定塊大小為8*4B （其中&為未命中，為命中）步長(zhǎng)命中情況1& 2& & 4& & & & 8& & & && & & &16& & & && & & &因?yàn)椴幻袝r(shí)間要遠(yuǎn)遠(yuǎn)大于命中時(shí)間，所以步長(zhǎng)為2時(shí)平均的運(yùn)行時(shí)間應(yīng)該是步長(zhǎng)1的2

2、倍，以此類推，到步長(zhǎng)為8時(shí)最接近2倍（注：假定塊中為8個(gè)字時(shí)），但步長(zhǎng)再增大時(shí)，就應(yīng)該不會(huì)再增加，或者增幅很小，代 碼如下:2)容量及層級(jí)數(shù)思路：不妨假定，cache是1KB的整數(shù)倍，可能值為1KB，2KB，4KB8MB(即代碼中的length )，由上一步可知塊大小為 64B，每塊有16字。每 次訪問(wèn)保證在1KB或者2KB或者4KB 的大小內(nèi)訪問(wèn)，假設(shè) L1大小 為16KB，當(dāng)剛超出范圍超出16KB時(shí)，訪問(wèn)時(shí)間會(huì)出現(xiàn)一個(gè)急劇的上升， 因?yàn)檫@時(shí)L1 cache大小已不滿足需要，必需借助L2，據(jù)此寫(xiě)如下代碼：1 #i nclude <stdio.h>2 #i nclude <t

3、ime.h>3 #defi ne PATH "/home/deropty/cache/cachesize.txt"4 #defi ne N 2*1024*1024/the maxsize is_5_2*1024*1024*4B=8MB6 int arrN;_7_int loop = (1<<24);8910 int main ()11 12 clock_t start, fini sh;13 int cou nt =1;14 FILE *fp = NULLA5-fp = fopen(PATH,"w");16 for (int lengt

4、h =256; length <= N; length +=256)1717 /256*4B=1KB,the step is18 1KB_20_-21/le ngth con trols the range_22_start = clock();23 int len gthmod = len gth -1;24 for (int i =0; i < loop; +i)25 +arr(i*16) & len gthmod;i*1626 to save the time27 28 fin ish = clock();29 printf( "%4dk: whe n th

5、e array len gth is %7d,30 the cost time is %7.3fms n ”,31 count,length,(double )(finish-start)32 / 1000);fprintf(fp, "%4dktt%7.3fmsn ", count,(double )(finish-start) /1000);+co unt;fclose(fp);fp = NULL;return 0;來(lái)自CODE的代碼片calCacheSize.c3)命中時(shí)間和缺失代價(jià)由1)已知塊大小為16字，所以每次訪問(wèn)16倍數(shù)個(gè)字，讓每次都缺失, 對(duì)大量缺失時(shí)間做均值

6、即可求出缺失代價(jià)。對(duì)于命中時(shí)間可一直訪問(wèn)同一 數(shù)據(jù)，對(duì)大量命中時(shí)間的數(shù)據(jù)做均值也可得出命中時(shí)間，代碼如下：1 #i nclude <stdio.h>2 #i nclude <time.h>_3 #i nclude <stdlib.h>45 #define N 1024*1024_6_int loop = (1<<25);78 int main ()9 10 int arrN;11 clock_t start, fini sh;12 double sumtime =0;43for (int k = 0; k <16; k +)14start

7、= clock();15for (int i =0; i < loop; +i)16int in dex = (i *16) & (N-1);17/gura ntee that it does miss18arri ndex *=3;/mea nin gless1920fin ish = clock();21int duration = finish - start;22start = clock();/mi nus23 the time taked by lopps24for (int i =0; i < loop; +i)25int in dex = (i*16) &a

8、mp; (N-1);2627fin ish = clock();28double tmp = ( double )(duration -29 (fin ish-start) / loop;30sumtime += tmp;31printf( "the miss time is %fusn ", tmp);printf("n The average miss time is%fusn ", sumtime16);sumtime=0;for (intk =0; k <16; k +)arr 0 =1;start = clock();for (int i

9、 =0; i < loop; +i)arr 0 *=3;/always visit42 1the arr0;)fin ish = clock();int duration = finish - start;start = clock();for (int i =0; i < loop; +i)484950fin ish = clock();51double tmp = ( double )(duration -52 (fin ish-start) / loop;53sumtime += tmp;Iprintf( "the hit time is %fusn ",

10、 tmp); printf("n The average hit timeis%fusn ”，sumtimej/16);return 0;來(lái)自CODE的代碼片calMissHitTime.c4 ）求緩存的關(guān)聯(lián)度思路：假設(shè)有每個(gè)緩存有16個(gè)塊，則直接映射情況如下：塊編號(hào)012345678910111213 1415塊二 012345678910111213 1415口 rh白出卜圭XT16171819202122232425 262728 29 3031映射情況32333435363738394041424344 45 4647484850515212路組相聯(lián)映射情況:|"塊

11、編號(hào)01234567塊 映射情況0123456789101112131415161718192021222324252627282930313233343536373839404142.4路組相聯(lián)映射情況塊編號(hào)0123塊 映射情況012345678910111213141516171819202122 232425262728293031323334 353637如上圖所示，對(duì)于編號(hào)為 0和16的塊，不停訪問(wèn)，在直接映射中會(huì)產(chǎn)生 沖突，但對(duì)2路、4路、8路則不會(huì)沖突；對(duì)于編號(hào)為0、16、32、48的塊，不停訪問(wèn)，在1路、2路中會(huì)產(chǎn)生沖突，但對(duì)于 4路、8路、16 路則不會(huì)沖突，以此類推，我們讓

12、程序每次都出現(xiàn)可能導(dǎo)致塊缺失的 情況，即訪問(wèn)編號(hào)為0、16的多次，計(jì)算平均訪問(wèn)時(shí)間；訪問(wèn)0、16、32、 48的塊多次，計(jì)算平均訪問(wèn)時(shí)間；訪問(wèn)編號(hào)為0、16、32、48、64、80的塊多次，計(jì)算平均訪問(wèn)時(shí)間 假定CPU為4路組相聯(lián)，則不停訪問(wèn)0、16、32、48不會(huì)出現(xiàn)缺失，而再增大為0、16、32、48、64，就會(huì)出現(xiàn)不停缺失的情況，據(jù)此思路，寫(xiě)出代碼如下：1 #i nclude <stdio.h>2 #in elude <time.h>丁4 #define N (1<<8<<10<<10)5 int arrN;6 int loop

13、 = (1<<16);/for producing a accurate result7-8 int main ()9 10 clock_t start, fini sh;11 int sumtime =0;1213for (intasso =2; asso <=16; asso +=2)14/suppose that15 associative is in (2-16)16start = clock();17for (int k =0; k < loop; +k)18for (int i =0; i < asso; +i)19arri<< 9<

14、<4 =0;20512blocks in cache L1, 16 in t/block21fin ish = clock();22int duration = finish - start;23start = clock();aga in for24 accuraty25for (int k =0; k < loop; +k)26for (int i =0; i < asso; +i)27i<< 9<<4<<4;28fin ish = clock();29printf( "%2dways associative's ave