求整数的二进制1的个数

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// Use this for initialization
void Start () {
//demo ();
//TimerManager.ShareInstance.OpenAllTimer ();

int num1 = fun1 ( 1000000 );
Debug.Log ( num1 );

int num2 = fun2 ( 1000000 );
Debug.Log ( num2 );

int num3 = fun3 ( 1000000 );
Debug.Log ( num3 );
}

/// <summary>
/// 求整数的二进制1的个数
/// </summary>
/// <param name="a"></param>
/// <returns></returns>
public int fun1 ( int a )
{
int num = 0;
while (a > 0)
{
a &= ( a - 1 );
num++;
}
return num;
}

/// <summary>
/// 求整数的二进制1的个数
/// </summary>
/// <param name="a"></param>
/// <returns></returns>
public int fun2 ( int a )
{
int num = 0;
while (a > 0)
{
num += a & 0x01;
a >>= 1;
}
return num;
}

/// <summary>
/// 求整数的二进制1的个数
/// </summary>
/// <param name="a"></param>
/// <returns></returns>
public int fun3 (int a)
{
int num = 0;
while (a > 0)
{
if (a % 2 == 1)
{
num++;
}
a = a / 2;
}
return num;
}

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