unsigned const STAT_ZERO = 0x0008;
unsigned const STAT_ONE = 0x0002;
unsigned const STAT_TWO = 0x0004;
unsigned const STAT_THREE = 0x0008;
// 检测位有1位是1就返回true
template <class T, class U>
bool isAnyBitSet(T value, U mask)
{
if ((value & mask) != 0)
return true;
else
return false;
}
// 检测位所有位都为1才返回true
template <class T, class U>
bool areAllBitsSet(T value, U mask)
{
if ((value & mask) == mask)
return true;
else
return false;
}
// 检测位全为0返回false
template <class T, class U>
bool areAllBitsClear(T value, U mask)
{
if ((value & mask) == 0)
return true;
else
return false;
}
// 置位
template <class T, class U>
T setBits(T value, U mask)
{
return (value | mask);
}
// 除了屏蔽位,其他全置位
template <class T, class U>
T setBitsExcept(T value, U mask)
{
return (value | ~mask);
}
// 清除指定位
template <class T, class U>
T clearBits(T value, U mask)
{
return (value & ~mask);
}
// 除了屏蔽位,其他全清0
template <class T, class U>
T clearBitsExcept(T value, U mask)
{
return (value & mask);
}
// n指定位置
template <class T>
T setBitByPos(T value, unsigned char n)
{
return (value | (1 << n));
}
template <class T>
T clearBitByPos(T value, unsigned char n)
{
return (value & ~(1 << n));
}
// 先置位add指定位,再清remove指定位为0
template <class T, class U>
T setBitByPos(T value, U add, U remove)
{
return ((value | add ) & ~remove);
}
template <class T>
bool isBitSetByPos(T value, unsigned char n)
{
if ((value & (1 << n)) != 0)
return true;
else
return false;
}
template <class T>
bool isBitClearByPos(T value, unsigned char n)
{
if ((value & (1 << n)) == 0)
return true;
else
return false;
}
// 获取二进制1个数
template <class T>
long getOneNum(T value)
{
int loop = sizeof(T)*8;
if (loop > sizeof(long)*8)
{
printf("value's bitwidth cannot longer than one word");
return 0;
}
int n=0;
int num = 0;
while (loop--)
{
if (isBitSetByPos(value, n++))
num++;
}
return num;
}
// 获取二进制0个数
template <class T>
long getZeroNum(T value)
{
int loop = sizeof(T)*8;
if (loop > sizeof(long)*8)
{
printf("value's bitwidth cannot longer than one word");
return 0;
}
int n=0;
int num = 0;
while (loop--)
{
if (isBitClearByPos(value, n++))
num++;
}
return num;
}
二进制处理函数
最新推荐文章于 2022-11-02 20:53:23 发布