求一个数的二进制位有多少个1

#include <stdio.h>
int main()
{
	int a = 13;
	//00000000000000000000000000001101
	//00000000000000000000000000000001
	int count = 0;
	int i = 0;
	for (i=0;i<32;i++)
	{
		if ((a & 1) == 1)
			count++;
		 a>>= 1;
	}
	printf("%d", count);
	return 0;
}

另外1:如果将13的二进制位中的从右向左第5位变为1

#include <stdio.h>
int main()
{
	int a = 13;
	//00000000000000000000000000001101
	//00000000000000000000000000010000
	//00000000000000000000000000011101
	int b = a | (1 << 4);
	printf("%d", b);
	return 0;
}

另外2：将29的二进制位中的从右向左第5位变回1

int main()
{
	int a = 29;
	//00000000000000000000000000011101
	//11111111111111111111111111101111
	//00000000000000000000000000001101
	int b = a & ~(1 << 4);
	printf("%d", b);
	return 0;
}

求一个数的二进制位有多少个1的另外的解法

//15 - 00001111
//15/2 - 7 - 0111
//7/2 - 3 - 0011
//3/2 - 1 - 0001
//1/2 - 0 - 0000
int Numberof1(unsigned int n)
{
	int count = 0;
	while (n)
	{
		if (n%2 == 1)
		{
			count++;
		}
		n /= 2;
	}
	return count;
}

int main()
{
	int n = 15;
	int ret = Numberof1(n);
	printf("%d", ret);
	return 0;
}

int Numberof1( int n)
{
	int count = 0;
	int i = 0;
	for (i = 0; i < 32; i++)
	{
		if (((n >> i) & 1) == 1)
		{
			count++;
		}
	}
	return count;
}

int main()
{
	int n = 15;
	int ret = Numberof1(n);
	printf("%d", ret);
	return 0;
}

//n=n&(n-1)
//n=15
//1111 - n
//1110 - n-1
//1110 - n
//1101 - n-1
//1100 - n
//1011 - n-1
//1000 - n
//0111 - n-1
//0000 - n
int Numberof1(int n)
{
	int count = 0;
	while (n)
	{
		n = n & (n - 1);
		count++;
	}
	return count;
}

int main()
{
	int n = 15;
	int ret = Numberof1(n);
	printf("%d", ret);
	return 0;
}

补充：写一个代码判断一个数是不是2的n次方

//写一个代码判断一个数是不是2的n次方
//2的n次方的数字，其二进制只有一个1
int main()
{
	int n = 0;
	scanf("%d", &n);
	if ((n & (n - 1)) == 0)
	{
		printf("是2的n次方");
	}
	else
	{
		printf("不是2的n次方");
	}
	return 0;
}