/// <summary>
/// Long 相乘(防止溢出操作)
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
/// <returns></returns>
public static long SaturatedMultiply(long a, long b)
{
// see checkedMultiply for explanation
int leadingZeros =
LongHelper.NumberOfLeadingZeros(a)
+ LongHelper.NumberOfLeadingZeros(~a)
+ LongHelper.NumberOfLeadingZeros(b)
+ LongHelper.NumberOfLeadingZeros(~b);
if (leadingZeros > 64 + 1)
{
return(a * b);
}
// the return value if we will overflow (which we calculate by overflowing a long :) )
long limit = long.MaxValue + LongHelper.UnsignedRightBitMove((a ^ b), (64 - 1));
if (leadingZeros < 64 | (a < 0 & b == long.MinValue))
{
// overflow
return(limit);
}
long result = a * b;
if (a == 0 || result / a == b)
{
return(result);
}
return(limit);
}
/// <summary>
/// Long bit的长度(Long有多少位为1)
/// 出处:java.lang.Long
/// </summary>
/// <param name="p_value"></param>
/// <returns></returns>
public static int BitCount(long p_value)
{
/*
* //Java 原始代码
* // HD, Figure 5-2
* i = i - ((i >>> 1) & 0x5555555555555555L);
* i = (i & 0x3333333333333333L) + ((i >>> 2) & 0x3333333333333333L);
* i = (i + (i >>> 4)) & 0x0f0f0f0f0f0f0f0fL;
* i = i + (i >>> 8);
* i = i + (i >>> 16);
* i = i + (i >>> 32);
* return (int)i & 0x7f;
*/
long i = p_value;
//ulong i = Convert.ToUInt64(p_value);
// HD, Figure 5-2
i = i - (LongHelper.UnsignedRightBitMove(i, 1) & 0x5555555555555555L);
i = (i & 0x3333333333333333L) + (LongHelper.UnsignedRightBitMove(i, 2) & 0x3333333333333333L);
i = (i + LongHelper.UnsignedRightBitMove(i, 4)) & 0x0f0f0f0f0f0f0f0fL;
i = i + LongHelper.UnsignedRightBitMove(i, 8);
i = i + LongHelper.UnsignedRightBitMove(i, 16);
i = i + LongHelper.UnsignedRightBitMove(i, 32);
return((int)i & 0x7f);
}
/// <summary>
/// 获取Long高位1的位置(左侧1的位置)
/// </summary>
/// <param name="i"></param>
/// <returns></returns>
public static int NumberOfTrailingZeros(long p_value)
{
/*
* //Java原始代码
* // HD, Figure 5-14
* int x, y;
* if (i == 0) return 64;
* int n = 63;
* y = (int)i; if (y != 0) { n = n -32; x = y; } else x = (int)(i>>>32);
* y = x <<16; if (y != 0) { n = n -16; x = y; }
* y = x << 8; if (y != 0) { n = n - 8; x = y; }
* y = x << 4; if (y != 0) { n = n - 4; x = y; }
* y = x << 2; if (y != 0) { n = n - 2; x = y; }
* return n - ((x << 1) >>> 31);
*/
long i = p_value;
// HD, Figure 5-14
int x, y;
if (i == 0)
{
return(64);
}
int n = 63;
y = (int)i; if (y != 0)
{
n = n - 32; x = y;
}
else
{
x = (int)LongHelper.UnsignedRightBitMove(i, 32);
}
y = x << 16; if (y != 0)
{
n = n - 16; x = y;
}
y = x << 8; if (y != 0)
{
n = n - 8; x = y;
}
y = x << 4; if (y != 0)
{
n = n - 4; x = y;
}
y = x << 2; if (y != 0)
{
n = n - 2; x = y;
}
return(n - IntHelper.UnsignedRightBitMove((x << 1), 31));
}
/// <summary>
/// Long 相减(防止溢出操作)
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
/// <returns></returns>
public static long SaturatedSubtract(long a, long b)
{
long naiveDifference = a - b;
if ((a ^ b) >= 0 | (a ^ naiveDifference) >= 0)
{
// If a and b have the same signs or a has the same sign as the result then there was no
// overflow, return.
return(naiveDifference);
}
// we did over/under flow
return(long.MaxValue + (LongHelper.UnsignedRightBitMove(naiveDifference, (64 - 1)) ^ 1));
}
/// <summary>
/// 该方法的作用是返回无符号整型i的最高非零位前面的0的个数，包括符号位在内；
///如果i为负数，这个方法将会返回0，符号位为1.
///比如说，10的二进制表示为 0000 0000 0000 0000 0000 0000 0000 1010
///java的整型长度为32位。那么这个方法返回的就是28
/// </summary>
/// <param name="i"></param>
/// <returns></returns>
public static int NumberOfLeadingZeros(long i)
{
/*
* // HD, Figure 5-6
* if (i == 0)
* return 64;
* int n = 1;
* int x = (int)(i >>> 32);
* if (x == 0) { n += 32; x = (int)i; }
* if (x >>> 16 == 0) { n += 16; x <<= 16; }
* if (x >>> 24 == 0) { n += 8; x <<= 8; }
* if (x >>> 28 == 0) { n += 4; x <<= 4; }
* if (x >>> 30 == 0) { n += 2; x <<= 2; }
* n -= x >>> 31;
* return n;
*/
// HD, Figure 5-6
if (i == 0)
{
return(64);
}
int n = 1;
int x = (int)LongHelper.UnsignedRightBitMove(i, 32);
if (x == 0)
{
n += 32; x = (int)i;
}
if (LongHelper.UnsignedRightBitMove(x, 16) == 0)
{
n += 16; x <<= 16;
}
if (LongHelper.UnsignedRightBitMove(x, 24) == 0)
{
n += 8; x <<= 8;
}
if (LongHelper.UnsignedRightBitMove(x, 28) == 0)
{
n += 4; x <<= 4;
}
if (LongHelper.UnsignedRightBitMove(x, 30) == 0)
{
n += 2; x <<= 2;
}
n -= IntHelper.UnsignedRightBitMove(x, 31);
return(n);
}
/// <summary>
/// Long 相加(防止溢出操作)
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
/// <returns></returns>
public static long SaturatedAdd(long a, long b)
{
/*
* long naiveSum = a + b;
* if ((a ^ b) < 0 | (a ^ naiveSum) >= 0) {
* // If a and b have different signs or a has the same sign as the result then there was no
* // overflow, return.
* return naiveSum;
* }
* // we did over/under flow, if the sign is negative we should return MAX otherwise MIN
* return Long.MAX_VALUE + ((naiveSum >>> (Long.SIZE - 1)) ^ 1);
*/
long naiveSum = a + b;
if ((a ^ b) < 0 | (a ^ naiveSum) >= 0)
{
// If a and b have different signs or a has the same sign as the result then there was no
// overflow, return.
return(naiveSum);
}
// we did over/under flow, if the sign is negative we should return MAX otherwise MIN
return(long.MaxValue + (LongHelper.UnsignedRightBitMove(naiveSum, (64 - 1)) ^ 1));
}