这是正常的求解方法,创建一个临时变量:
int mian()
{
int a = 3;
int b = 5;
int tmp = 0;
printf("before: a = %d b = %d\n", a, b);
tmp = a;
a = b;
b = tmp;
printf("before: a = %d b = %d\n", a, b);
return 0;
}
在不创建临时变量的情况下交换两个数的数值:
①加减法
int main()
{
int a = 3;
int b = 5;
a = a + b; //8
b = a - b; //3
a = a - b;// 5
printf("a = %d,b = %d\n", a, b);
return 0;
}
但是你会发现这里是有缺陷的,因为当你的a的值或者b的值足够大,但是没有超过整形的最大值时,两个数相加就会超多整形的最大范围,会出现溢出,那样就会报错,所以方法还需改进。
②异或方法
int main()
{
int a = 3;
int b = 5;
a = a^b;
b = a^b;
a = a^b;
printf("a = %d b = %d\n", a, b);
return 0;
}
求一个整数存储在内存中的二进制中1的个数(统计num的补码中有几个1)
int main()
{
int num = 0;
int count = 0;
scanf("%d", &num);
while (num)
{
if (num % 2 == 1)
{
count++;
num = num / 2;
}
}
printf("%d ", count);
}
//但是你会发现当你输入的值是-1的时候他的补码本应该是32个1,但是结果显示是0,这里就报错了
按位&的方法求解其中有多少1
int main()
{
int num = 0;
int i = 0;
int count = 0;
scanf("%d", &num);
for (i = 0; i < 32; i++)
{
if (1 == (num >> i) & 1)
{
count++;
}
}
printf("%d ", count);
return 0;
}
#include<stdio.h>
//这里还有一种更加NB的思想方式
//13
//1101 n
//1100 n-1
//------------
//1100 n
//1011 n-1
//------------
//1000 n
//0111 n-1
//-----------
//0000
//每&一次就会消除掉最右边的一个1
int count_bit_one(int n)
{
int count = 0;
while (n)
{
n = n&(n - 1);
count++;
}
return count;
}
int main()
{
int num = 0;
scanf("%d",num);
int count = 0;
count = count_bit_one(num);
printf("count = %d",count);
return 0;
}
编程实现:两个int(32位)整数m和n的二进制表达中,有多少个位(bit)不同?
输入例子 : 1999 2299
输出例子 : 7
#include<stdio.h>
int main()
{
int m = 0;
int n = 0;
int ret = 0;
int i = 0;
int count = 0;
scanf("%d%d", &m, &n);
ret = m^n;
for (i = 0; i < 32; i++)
{
if (1 == ((ret >> i) & 1))
{
count++;
}
}
printf("count = %d", count);
return 0;
}
②函数的方式封装来实现
#include<stdio.h>
int count_bit_one(int n)
{
int count = 0;
while (n)
{
n = n&(n - 1);
count++;
}
return count;
}
int get_diff_bit(int m, int n)
{
int tmp = m^n;
return count_bit_one(tmp);
}
int main()
{
int m = 0;
int n = 0;
scanf("%d%d", &m, &n);
int count = 0;
count = get_diff_bit(m, n);
printf("count = %d", count);
return 0;
}
获取一个整数二进制序列中所有的偶数位和奇数位,分别打印出二进制序列
void Print(int m)
{
int i = 0;
printf("奇数位\n");
for (i = 30; i >= 0; i -= 2)
{
printf("%d ", ((m >> i) & 1));
}
printf("\n");
printf("偶数位\n");
for (i = 31; i >= 1; i -= 2)
{
printf("%d ", ((m >> i) & 1));
}
}
int main()
{
int m = 0;
scanf("%d", &m);
Print(m);
return 0;
}