Count the number of 1 in the binary (four scenarios)

Scenario One: (only suitable for calculating positive numbers)

#include <stdio.h>

#include <stdlib.h>

int main ()

{

int num = 10; 10 binary number is 1010

int count = 0;

while (NUM)

{

if (num% 2 = = 1)//start with the highest bit, the remainder is 1 is 1, the remainder is 2 is 0

{

count++; Remainder is 1 o'clock count plus 1

}

num = NUM/2; Dividing by 2 is equivalent to moving one bit to the right, i.e. losing the counted

}

printf ("%d\n", count);

System ("pause");

return 0;

}

Scenario Two: (positive negative number can be)

#include <stdio.h>

#include <stdlib.h>

int main ()

{

int num = 10;

int count = 0;

while (NUM)

{

if ((num&1) = = 1)//& bitwise AND: There are 0 0, and the double 1 is 1. That is, starting from the highest bit, each digit is bitwise AND 1.

{

count++;

}

num = num >> 1; Equivalent to dividing by 2.

}

printf ("%d\n", count);

System ("pause");

return 0;

}

Scenario Three: (positive negative number can be)

#include <stdio.h>

#include <stdlib.h>

int main ()

{

unsigned int num =-1; -1 binary number is 11111111 11111111 11111111 11111111

int count = 0;

while (NUM)

{

if ((num&1) = = 1)

{

count++;

}

num = num >> 1;

}

printf ("%d\n", count);

System ("pause");

return 0;

}

Scheme four: (positive negative number can be)

#include <stdio.h>

#include <stdlib.h>

int main ()

{

int num =-1;

int count = 0;

while (NUM)

{

count++;

num = num& (num-1);

}

printf ("%d\n", count);

System ("pause");

return 0;

}

Results:

The binary number of num=10//10 is 1010

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Num=-1//-1 binary number is 11111111 11111111 11111111 11111111

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Count the number of 1 in the binary (four scenarios)