C/C++/C#/JAVA/單晶片 位元運算 應用

資料來源: https://blog.kuoe0.tw/posts/2012/01/26/bitwise-operation-utlization/

01.奇數/偶數 判斷

if((x & 1) == 1)
{

}
else
{

}

02.利用左移運算，即可輕易將整數 x 乘以 2^n

x << n

03.利用右移運算，即可輕易將整數 x 除以 2^n

x >> n

04.末端連續 n 個位元 1

(1 << n) - 1

05.取 2^n 的模數

x & ((1 << n) - 1)

06.最低的位元 1

x & -x

07.是否為二的冪次(2的次方)

if((x & -x) == x)
{
}
else
{
}

08.將第 n 個位元設定為 0

x & ~(1 << n)

09.整數交換(SWAP)

x^=y^=x^=y

10.消去最低位 1 位元

x & ( x - 1 )

11.將第 n 個位元設定為 1

x | ( 1 << n )

測試範例

#include <stdio.h>
#include <stdlib.h>

void pause()
{
    printf("Press Enter key to continue...");
    fgetc(stdin);
}

int sky_itoa(int value, char *str, unsigned int radix)
{
	char list[] = "0123456789ABCDEF";
	unsigned int tmp_value;
	int i, j, k;

	if (NULL == str)
	{
		return 1;
	}

	if (2 != radix && 8 != radix && 10 != radix && 16 != radix)
	{
		return 1;
	}
		i = 0;
		k = 0;
		if (radix == 10 && value < 0) {
		tmp_value = (unsigned int)(0 - value);
		str[i++] = '-';
		k = 1;
	}
	else
	{
		tmp_value = (unsigned int)value;
	}

	do
	{
		str[i++] = list[tmp_value%radix];
		tmp_value /= radix;
	} while(tmp_value);
	str[i] = '\0';

	//翻转
	char tmp;
	for (j = k; j < (i+k)/2; j++)
	{
		tmp = str[j];
		str[j] = str[i+k-j-1];
		str[i+k-j-1] = tmp;
	}
	return 0;
}

int main()
{
    int x=10;
    char str[100] ={0};
    sky_itoa(x,str,2);
    printf("%d=>%s\n",x,str);

    str[0] ='\0';
    if((x & 1) == 1)
    {
        printf("%d=>odd\n",x);
    }
    else
    {
        printf("%d=>even\n",x);
    }

    printf("%d*(2^3)=>%d\n",x,(x << 3));

    printf("%d/(2^3)=>%d\n",x,(x >> 3));

    str[0] ='\0';
    sky_itoa((1 << 10) - 1,str,2);
    printf("%s\n",str);

    printf("%d\n",( x & ((1 << 3) - 1) ) );

    printf("%d\n",(x & -x));

    if((8 & -8) == 8)
    //if((x & -x) == x)
    {
        printf("%c\n",'O');
    }
    else
    {
        printf("%c\n",'X');
    }

    printf("%d\n",(x & ~(1 << 3)));

    printf("%d\n",(x | (1 << 2)));
    pause();
    return 0;
}