TCPL/2.09_Bitwise_Operators

2.9 Bitwise Operators 按位运算符

C provides six operators for bit manipulation; these may only be applied to integral operands, that is, char, short, int, and long, whether signed or unsigned.

&       bitwise AND
|       bitwise inclusive OR
^       bitwise exclusive OR
<<      left shift
>>      right shift
~       one's complement (unary)

C语言提供了6个位操作运算符。这些运算符只能作用于整型操作数，即只能作用于带符号或无符号的char、short、int与long类型：

&       按位与(AND)
|       按位或(OR)
^       按位异或(XOR)
<<      左移
>>      右移
~       按位求反(一元运算符)

The bitwise AND operator & is often used to mask off some set of bits, for example

   n = n & 0177;

sets to zero all but the low-order 7 bits of n.

按位与运算符&经常用于屏蔽某些二进制位，例如：

   n = n & 0177;

该语句将n中除7个低二进制位外的其他各位均置为0。

The bitwise OR operator | is used to turn bits on:

   x = x | SET_ON;

sets to one in x the bits that are set to one in SET_ON.

按位或运算符常用于将某些二进制位置为1，例如：

   x = x | SET_ON;

该语句将x中对应于SET_ON中为1的那些二进制位置为1；

The bitwise exclusive OR operator ^ sets a one in each bit position where its operands have different bits, and zero where they are the same.

按位异或运算符^当两个操作数的对应位不相同时将该位设置为1，否则，将该位设置为0。

One must distinguish the bitwise operators & and | from the logical operators && and ||, which imply left-to-right evaluation of a truth value. For example, if x is 1 and y is 2, then x & y is zero while x && y is one.

我们必须将位运算符&、|同逻辑运算符&&、||区分开来，后者用于从左至右求表达式的真值。例如，如果x的值为1，y的值为2，那么，x&y的结果为0，而x&&y的值为1。

The shift operators << and >> perform left and right shifts of their left operand by the number of bit positions given by the right operand, which must be non-negative. Thus x << 2 shifts the value of x by two positions, filling vacated bits with zero; this is equivalent to multiplication by 4. Right shifting an unsigned quantity always fits the vacated bits with zero. Right shifting a signed quantity will fill with bit signs ("arithmetic shift") on some machines and with 0-bits ("logical shift") on others.

移位运算符<<与>>分别用于将运算的左操作数左移与右移，移动的位数则由右操作数指定(右操作数的值必须是非负值)。因此，表达式x<<2将把x的值左移2位，右边空出的2位用0填补，该表达式等价于对左操作数乘以4。在对unsigned类型的无符号值进行右移位时，左边空出的部分将用0填补；当对signed类型的带符号值进行右移时，某些机器将对左边空出的部分用符号位填补(即“算术移位”)，而另一些机器则对左边空出的部分用0填补(即“逻辑移位”)。

The unary operator ~ yields the one's complement of an integer; that is, it converts each 1-bit into a 0-bit and vice versa. For example

   x = x & ~077

sets the last six bits of x to zero. Note that x & ~077 is independent of word length, and is thus preferable to, for example, x & 0177700, which assumes that x is a 16-bit quantity. The portable form involves no extra cost, since ~077 is a constant expression that can be evaluated at compile time.

一元运算符~用于求整数的二进制反码，即分别将操作数各二进制位上的1变为0，0变为1。例如：

   x = x & ~077

将把x的最后6位设置为0。注意，表达式x&~077与机器字长无关，它比形式为x&0l77700的表达式要好，因为后者假定x是16位的数值。这种可移植的形式并没行增加额外开销，因为~077是常量表达式，可以在编译时求值。

As an illustration of some of the bit operators, consider the function getbits(x,p,n) that returns the (right adjusted) n-bit field of x that begins at position p. We assume that bit position 0 is at the right end and that n and p are sensible positive values. For example, getbits(x,4,3) returns the three bits in positions 4, 3 and 2, right-adjusted.

   1    /* getbits:  get n bits from position p */
   2    unsigned getbits(unsigned x, int p, int n)
   3    {
   4        return (x >> (p+1-n)) & ~(~0 << n);
   5    }

The expression x >> (p+1-n) moves the desired field to the right end of the word. ~0 is all 1-bits; shifting it left n positions with ~0<<n places zeros in the rightmost n bits; complementing that with ~ makes a mask with ones in the rightmost n bits.

为了进一步说明某些位运算符，我们来看函数getbits(x,p,n)，它返回x中从右边数第p位开始向右数n位的字段。这里假定最右边的一位是第0位，n与p都是合理的正值。例如，getbits(x,4,3)返回x中第4、3、2三位的值。

   1    /* getbits:  get n bits from position p */
   2    unsigned getbits(unsigned x, int p, int n)
   3    {
   4        return (x >> (p+1-n)) & ~(~0 << n);
   5    }

其中，表达式x >> (p+1-n)将期望获得的字段移位到字的最右端。~0的所有位都为1，这里使用语句~0<<n将~0左移n位，并将最右边的n位用0填补。再使用~运算对它按位取反，这样就建立了最右边n位全为1的屏蔽码。

Exercise 2-6. Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.

练习2-6 编写一个函数setbits(x,p,n,y)，该函数返回对x执行下列操作后的结果值：将x中从第p位开始的n个(二进制)位设置为y中最右边n位的值，x的其余各位保持不变。

Exercise 2-7. Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.

练习2-7 编写一个函数invert(x,,p,n)，该函数返回对x执行下列操作后的结果值：将x中从第p位开始的n个(二进制)位求反(即，1变为0，0变成1)，x的其余各位保持不变。

Exercise 2-8. Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n positions.

练习2-8 编写一个函数rightrot(x,n)，该函数返回将x循环右移(即从最右端移出的位将从最左端移入)n(二进制)位后所得到的值。