数据压缩位操作中的一个函数bit_rot_left "位向左轮转"看不懂,望指点

Question

下面代码中bitget和bitset是bitrotleft中要用到的简单函数,功能分别是取得和设置指定位的bit值,我很难看懂bitrotleft函数,虽然书上简介写的很简单:轮转缓冲区bits(含size位),将位值向左移count位.此操作完成后,处于最左端的count位移动到缓冲区最右端,而且其他的位也相应的轮转.但是我在main函数的测试中好像并不如意,代码如下:#include#includeintbit_get(constunsignedchar*bits,intpos){unsignedcharmask;inti;/*设置掩码*/mask=0x80;for(i=0;i>1;/*获得当前位的数值*/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

Answer 1

测试程序应该这样写才对:
intmain()
{
unsignedcharbits[3]={0x4b,0xdb,0};
bit_rot_left(bits,16,2);
printf("%#x,%#X,%#x",bits[0],bits[1],bits[2]);
}
输出是
0x2f,0x6D,0
                            

Answer 2

嗯你那个程序有错。。。我没有去调试它，不过我写了一个好用的：
#include
intgcd(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;
}
输出如下
0011000100110000001100000011000100110000001100010011000000110001
????????
1100010011000000110000001100010011000000110001001100000011000100