C:位运算

本文转自：http://zhangjunhd.blog.51cto.com/113473/52734

1.负数表示-二类补数(twos complement)
一般使用二类补数表示负数,最左边一位为符号位.
将一个十进制负数转化为二进制符号数时,首先给这个数加1,然后取绝对值,再将其转换为二进制,最后对这个二进制求补.
[1]-5
[2]-4
[3]4
[4]0000 0100
[5]1111 1011
将一个负数从二进制转换为十进制,首先对其所有位求补,然后将结果转换为十进制,再改变其符号,最后再减1.
[1]1111 1011
[2]0000 0100
[3]4
[4]-4
[5]-5
使用二类补数表示数字时,用n位可存储的最大正数是2^n-1-1,用n位可存储的最小负数-2^n-1.
假设整数占32位(4字节),此时可存储的最大正数2³¹-1,最小负数-2³¹,而如果是无符号整数,即unsigned int,其表示范围为0至2³²-1.

2.按位与和按位或
按位与经常用于屏蔽一个数中的某些位,
word &= 0x1//除最右边4位外,其余位清零
按位或经常用于将某些位设定为1
word |= 0x1//将最右边4位设定为1

3.异或运算
异或运算可以交换两个值而不需要使用临时变量.

/*exchange two integer values*/
void swap(int* a, int* b)
{
 *a ^= *b;
 *b ^= *a;
 *a ^= *b;
}

如果某数与1进行异或运算,可以达到取反的效果,0^1=1,1^1=0.
这里假设机器的整型数长度为32位,对整数0的31位取反,得到最大整数.

int main()
{
 int a=0;
 printf("%i,%i",a,a ^ 0x7fffffff);
 return 0;
}

4.几个经典的位运算函数
4.1求当前机器无符号整型最大长度

/*the max length of unsigned int*/
int int_size ()
{
 unsigned int bits;
 int size = 0;

 bits = ~0;

 while ( bits ) {
 ++size;
 bits >>= 1;
 }
 return size;
}

4.2移位运算

/*a bit mover for unsigned int
 if n > 0 move left for n bits,else move right*/
unsigned int bit_shift (unsigned int value,int n)
{
 int intsize=int_size(); /*the length of unsigned int*/

 if(n>0 && n< intsize) /*move left*/
 value<<=n;
 else if (n<0 && n> -intsize) /*move right*/
 value>>=-n;
 else
 value=0;
 return value;
}

4.3循环移位运算

/*a bit rotate mover for unsigned int
 if n > 0 move left for n bits,else move right*/
unsigned int bit_rotate (unsigned int value, int n)
{
 unsigned int result,bits,intsize;

 intsize=int_size(); /*the length of unsigned int*/

 if(n > 0)
 n=n % intsize;
 else
 n=-(-n % intsize);

 if(n==0)
 result=value;
 else if(n >0 ){ /*move left*/
 bits=value >> (intsize-n);/*bits should be in the rightest*/
 result=value << n|bits;
 }else{ /*move right*/
 n=-n;
 bits=value << (intsize-n);/*bits should be in the leftest*/
 result=value >> n|bits;
 }
 return result;
}

4.4返回无符号整型数value中从右起第p位的值

/* get bit No.p(from right) of value to see if it is on */
int bit_get (unsigned int value, int n)
{
 int intsize=int_size(); /*the length of unsigned int*/

 if ( p < 0 || p > intsize-1 )/*out of range*/
 return 0;

 if ( (value >> p) & 1 )
 return 1;
 else
 return 0;
}

4.5将无符号整型数value中从右起的第p位置1

/* set bit No.p(from right) of value on */
unsigned int bit_set (unsigned int value, int p)
{
 int intsize=int_size(); /*the length of unsigned int*/

 if ( p < 0 || p > intsize-1 )/*out of range*/
 return 0;

 return value | (1 << p);
}

4.6返回无符号整型数value中从第p位(右起)向右n位的值
[1]~(~0 << n)表示最右边n位全为1;
[2]value >> (p+1-n)表示将目标位字段移至最右端;

/*get n bits of value at position p(from right) */
unsigned bits_get (unsigned int value, int p, int n)
{
 int intsize=int_size(); /*the length of unsigned int*/

 if ( n < 0 || p < 0 || p + n > intsize )
 return 0;
 return(value >> (p+1-n)) & ~(~0 << n);
}

4.7将无符号整型数value中从第p位(右起)向右n位设置为y最右边n位的值
[1]~(~0 << n)表示最右边n位全为1;
[2](~(~0 << n) << (p+1-n)表示将这n个1位左移至位置p;
[3]~(~(~0 << n) << (p+1-n))表示将从位置p开始的n位设置零,其余位设置一;
[4]unsigned tar=bits_get(y,n-1,n);取出y的低n位;

/*set n bits of value at position p(from right) with bits of y*/
unsigned bits_set (unsigned value, int p, int n, unsigned int y)
{
 int intsize=int_size(); /*the length of unsigned int*/

 if ( n < 0 || p < 0 || p + n > intsize )
 return 0;
 unsigned tar=bits_get(y,n-1,n);
 return (value & ~(~(~0 << n) << (p+1-n))) | tar;
}