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數(shù)據(jù)壓縮位操作中的一個(gè)函數(shù)bit_rot_left "位向左輪轉(zhuǎn)"看不懂,望指點(diǎn)

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蕪湖不蕪 2019-04-13 08:45:42

下面代碼中bitget和bitset是bitrotleft中要用到的簡(jiǎn)單函數(shù),功能分別是取得和設(shè)置指定位的bit值,我很難看懂bitrotleft函數(shù),雖然書上簡(jiǎn)介寫的很簡(jiǎn)單:輪轉(zhuǎn)緩沖區(qū)bits(含size位),將位值向左移count位.此操作完成后,處于最左端的count位移動(dòng)到緩沖區(qū)最右端,而且其他的位也相應(yīng)的輪轉(zhuǎn).但是我在main函數(shù)的測(cè)試中好像并不如意,代碼如下:#include#includeintbit_get(constunsignedchar*bits,intpos){unsignedcharmask;inti;/*設(shè)置掩碼*/mask=0x80;for(i=0;i>1;/*獲得當(dāng)前位的數(shù)值*/return(((mask&bits[(int)(pos/8)])==mask)?1:0);}voidbit_set(unsignedchar*bits,intpos,intstate){unsignedmask;inti;mask=0x80;for(i=0;i>1;if(state)bits[pos/8]|=mask;elsebits[pos/8]&=(~mask);return;}voidbit_rot_left(unsignedchar*bits,intsize,intcount){intfbit,lbit,i,j;/*Rotatethebuffertotheleftthespecifiednumberofbits.*/if(size>0){for(j=0;j

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德瑪西亞99

TA貢獻(xiàn)1770條經(jīng)驗(yàn) 獲得超3個(gè)贊

測(cè)試程序應(yīng)該這樣寫才對(duì):intmain(){unsignedcharbits[3]={0x4b,0xdb,0};bit_rot_left(bits,16,2);printf("%#x,%#X,%#x",bits[0],bits[1],bits[2]);}輸出是0x2f,0x6D,0

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慕村9548890

TA貢獻(xiàn)1884條經(jīng)驗(yàn) 獲得超4個(gè)贊

嗯你那個(gè)程序有錯(cuò)。。。我沒有去調(diào)試它，不過我寫了一個(gè)好用的：#includeintgcd(inta,intb){returnb==0?a:gcd(b,a%b);}voidswap(unsignedchar*a,unsignedchar*b){unsignedchartmp=*a;*a=*b;*b=tmp;}voidrot1(unsignedchar*bits,intsize,intcount){intm=gcd(size,count),i,j;unsignedchartmp;count%=size;count=size-count;for(i=0;i{tmp=bits[i];for(j=i+count;j!=i;j+=count){if(j>size)j-=size;swap(&tmp,bits+j);}bits[i]=tmp;}}voidrot2(unsignedchar*bits,intsize,intcount){inti;unsignedcharmask=~0<<(8-count),tmp1=0,tmp2;for(i=size-1;i>=0;i--){tmp2=(mask&bits[i])>>(8-count);bits[i]<<=count;bits[i]|=tmp1;tmp1=tmp2;}bits[size-1]|=tmp1;}voidbit_rot_left(unsignedchar*bits,intsize,intcount){rot1(bits,size,count/8);rot2(bits,size,count%8);}voidprint_bits(unsignedchar*bits,intsize){inti,j;for(i=0;ifor(j=0x80;j;j>>=1)printf("%d",!!(j&bits[i]));printf("\n");}intmain(){unsignedcharbits[]="10010101";print_bits(bits,8);bit_rot_left(bits,8,2);printf("%s\n",bits);print_bits(bits,8);return0;}輸出如下0011000100110000001100000011000100110000001100010011000000110001????????1100010011000000110000001100010011000000110001001100000011000100