int main()
{
int a=2,b=0;
int i=0;
for(;i<32;i++)
{
b+=((a>>i)&1)<
}
printf(“%d\n”,b);
return 0;
}
晕刚回答了；还没有关
http://topic.csdn.net/u/20080504/10/9579d8d9-895a-492c-b597-4b3de3057058.html?seed=1285008557
//这是位翻转的
//互换是同样的技巧
#include
voidShowBit(unsigned int x, int n)
{
if (–n) ShowBit(x>>1, n);
printf(“%d”, x%2);
}
voidReverseBit(unsigned int* pValue)
{
unsigned int n = *pValue;
n = (n&0x55555555)<<1|(n&0xAAAAAAAA)>>1;
n = (n&0x33333333)<<2|(n&0xCCCCCCCC)>>2;
n = (n&0x0F0F0F0F)<<4|(n&0xF0F0F0F0)>>4;
n = (n&0x00FF00FF)<<8|(n&0xFF00FF00)>>8;
n = (n&0x0000FFFF)<<16|(n&0xFFFF0000)>>16;
*pValue = n;
}
int main()
{
unsigned int x = 0x6A;
ShowBit(x, 32); printf(“\n”);
ReverseBit(&x);
ShowBit(x, 32); printf(“\n”);
return 0;
}
楼上的方法参考<> 一般的方法可以这样来：
int reverse(unsigned x)
{
int m = 0, i;
unsigned mask = x;
for (i=0; i<32; i++)
{
m = (m <
mask >>= 1;
}
return m;
}
int main(int argc, char* argv[])
{
int a=1;
int highLow;
int lowHigh;
int result;
highLow=a>>16;//右移(舍弃右边的16位,左边高位移入0),实现了高位变低位
lowHigh=a<<16;//左移(舍弃左边的16位,右补0),实现了低位变高位
result=highLow+lowHigh;//相加就是结果拉,,,
cout<
return 0;
}
int main(int argc, char* argv[])
{
int a=1;
int highLow;
int lowHigh;
int result;
highLow=a>>16;//右移(舍弃右边的16位,左边高位移入0),实现了高位变低位
lowHigh=a
result=highLow+lowHigh;//相加就是结果拉,,,
cout
return 0;
}
ls正解，很好理解。
问题有歧异，lz最好举例说明下你的要求
楼主开了两个贴啊
up!
大家提出这么多方法了啊！
我再贴出一个，呵呵！
#include
using std::cout;
using std::endl;
using std::hex;
int reverseInt(int);
int main()
{
int a = 0x1fed1234;
cout<
int b = reverseInt(a);
cout<
system(“pause”);
}
int reverseInt(int a)
{
return ((a&0xffff0000) >>16) | ((a&0x0000ffff) <<16);
}
如果是位翻转，就看飞雪的了
左移右移就好了
UINT _b = 0xF2345678;
_asm
{
mov eax, _b
shl eax, 16
mov edx, _b
shr edx, 16
or eax, edx
mov _b, eax
}
printf(“0x%x\n”, _b);