给定一个整数typedef：
typedef unsigned int TYPE;
要么
typedef unsigned long TYPE;
我有以下代码来反转整数的位：
TYPE max_bit= (TYPE)-1;
void reverse_int_setup()
{
TYPE bits= (TYPE)max_bit;
while (bits <<= 1)
max_bit= bits;
}
TYPE reverse_int(TYPE arg)
{
TYPE bit_setter= 1,bit_tester= max_bit,result= 0;
for (result= 0; bit_tester; bit_tester>>= 1,bit_setter<<= 1)
if (arg & bit_tester)
result|= bit_setter;
return result;
}
首先需要运行reverse_int_setup(),它存储一个打开最高位的整数,然后对reverse_int(arg)的任何调用返回arg,其位反转(用作二叉树的一个键,取自一个增加反击,但这或多或少无关紧要).
在调用reverse_int_setup()之后,是否存在一种与平台无关的方法在编译时为max_int提供正确的值;否则,是否有一个算法比你对reverse_int()更好/更精简？
谢谢.
解决方法
#include<stdio.h> #include<limits.h> #define TYPE_BITS sizeof(TYPE)*CHAR_BIT typedef unsigned long TYPE; TYPE reverser(TYPE n) { TYPE nrev = 0,i,bit1,bit2; int count; for(i = 0; i < TYPE_BITS; i += 2) { /*In each iteration,we swap one bit on the 'right half' of the number with another on the left half*/ count = TYPE_BITS - i - 1; /*this is used to find how many positions to the left (and right) we gotta move the bits in this iteration*/ bit1 = n & (1<<(i/2)); /*Extract 'right half' bit*/ bit1 <<= count; /*Shift it to where it belongs*/ bit2 = n & 1<<((i/2) + count); /*Find the 'left half' bit*/ bit2 >>= count; /*Place that bit in bit1's original position*/ nrev |= bit1; /*Now add the bits to the reversal result*/ nrev |= bit2; } return nrev; } int main() { TYPE n = 6; printf("%lu",reverser(n)); return 0; }
这次我使用了来自TK的’位数’的想法,但是假设一个字节包含8位而不是使用CHAR_BIT宏,它使它更具可移植性.现在代码效率更高(删除了内部for循环).我希望这次代码也略显神秘.
