C:位運算

本文轉自：http://zhangjunhd.blog.51cto.com/113473/52734

1.負數表示-二類補數(twos complement)
一般使用二類補數表示負數,最左邊一位爲符號位.
將一個十進制負數轉化爲二進制符號數時,首先給這個數加1,然後取絕對值,再將其轉換爲二進制,最後對這個二進制求補.
[1]-5
[2]-4
[3]4
[4]0000 0100
[5]1111 1011
將一個負數從二進制轉換爲十進制,首先對其所有位求補,然後將結果轉換爲十進制,再改變其符號,最後再減1.
[1]1111 1011
[2]0000 0100
[3]4
[4]-4
[5]-5
使用二類補數表示數字時,用n位可存儲的最大正數是2^n-1-1,用n位可存儲的最小負數-2^n-1.
假設整數佔32位(4字節),此時可存儲的最大正數2³¹-1,最小負數-2³¹,而如果是無符號整數,即unsigned int,其表示範圍爲0至2³²-1.

2.按位與和按位或
按位與經常用於屏蔽一個數中的某些位,
word &= 0x1//除最右邊4位外,其餘位清零
按位或經常用於將某些位設定爲1
word |= 0x1//將最右邊4位設定爲1

3.異或運算
異或運算可以交換兩個值而不需要使用臨時變量.

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

如果某數與1進行異或運算,可以達到取反的效果,0^1=1,1^1=0.
這裏假設機器的整型數長度爲32位,對整數0的31位取反,得到最大整數.

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

4.幾個經典的位運算函數
4.1求當前機器無符號整型最大長度

/*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移位運算

/*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循環移位運算

/*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返回無符號整型數value中從右起第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將無符號整型數value中從右起的第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返回無符號整型數value中從第p位(右起)向右n位的值
[1]~(~0 << n)表示最右邊n位全爲1;
[2]value >> (p+1-n)表示將目標位字段移至最右端;

/*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將無符號整型數value中從第p位(右起)向右n位設置爲y最右邊n位的值
[1]~(~0 << n)表示最右邊n位全爲1;
[2](~(~0 << n) << (p+1-n)表示將這n個1位左移至位置p;
[3]~(~(~0 << n) << (p+1-n))表示將從位置p開始的n位設置零,其餘位設置一;
[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;
}