C language: Find the number of different bits in the binary of two numbers

topic:

Programming implementation : How many bits (bits) are different in the binary representation of two int (32-bit) integers m and n ?

              

Input example:

1999 2299

               

Example output: 7

                    

 =========================================================================

                       

Ideas:

general idea:

After XORing m and n , there are as many 1s as there are differences , and then calculate how many 1s there are in binary

          

(1). Write a function: number_of_1() to calculate how many 1s there are in binary .

               

(2). In the main function , pass the result of m ^ n (m XOR n) as a parameter to the number_of_1() function

                

---

                 

first step:

Write a function: number_of_1() to calculate how many 1s there are in binary :

           

(1). Function return value: int , returns the number of 1 in binary ;

Function parameter: int m

                   

(2). Set the counter count to count the number of 1 in the two-level system

             

(3). Use while loop , loop statistics, continue to loop as long as m is not 0

              

(4). Use m = m & (m - 1); --Remove the rightmost 1 in the binary, and then assign it to m itself ,

Remove a 1 , and the counter count is ++ .

                     

Implementation code:

#include <stdio.h>

//编写函数：
int number_of_1(int m)
//函数返回值：int ； 函数形参：int m
{
	//设置计数器：
	int count = 0;

	//使用while循环
	while (m) 
	//如果m不为0，就说明二进制中还有1，则继续循环
	{
		//使用公式：
		m = m & (m - 1); //去掉最右边的1
		count++; //计数器++
	}
}

int main()
{
	
	return 0;
}

Realize the picture:

                 

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Step two:

main function:

                 

(1). Input two parameters : m and n

             

(2). Pass the result of m ^ n (m XOR n) as a parameter to the number_of_1() function ,

and use a variable to receive the return value

                 

(3). Print return value

                     

Implementation code:

#include <stdio.h>

//编写函数：
int number_of_1(int m)
//函数返回值：int ； 函数形参：int m
{
	//设置计数器：
	int count = 0;

	//使用while循环
	while (m) 
	//如果m不为0，就说明二进制中还有1，则继续循环
	{
		//使用公式：
		m = m & (m - 1); //去掉最右边的1
		count++; //计数器++
	}
}

int main()
{
	//输入两个数：
	int m = 0;
	int n = 0;
	scanf("%d %d", &m, &n);

	//异或：相同为0，相异为1
	int ret = number_of_1(m ^ n);

	//打印：
	printf("%d", ret);

	return 0;
}

Realize the picture:

                    

Final code and implementation effect

Final code:

#include <stdio.h>

//编写函数：
int number_of_1(int m)
//函数返回值：int ； 函数形参：int m
{
	//设置计数器：
	int count = 0;

	//使用while循环
	while (m) 
	//如果m不为0，就说明二进制中还有1，则继续循环
	{
		//使用公式：
		m = m & (m - 1); //去掉最右边的1
		count++; //计数器++
	}
}

int main()
{
	//输入两个数：
	int m = 0;
	int n = 0;
	scanf("%d %d", &m, &n);

	//异或：相同为0，相异为1
	int ret = number_of_1(m ^ n);

	//打印：
	printf("%d", ret);

	return 0;
}

Realize the effect: