Idea 1: Judging by the method of remainder
#include<stdio.h>
intmain()
{
int a=0;
int i=0,count=0;
printf("Please enter a positive integer: ");
scanf("%d",&a);
do
{
if((a%2)==1)
{
count++;
a/=2;
}
}while(a);
printf("Number of 1s in binary in this positive integer:%d\n",count);
return 0;
}
Defect: Negative numbers cannot be processed. Negative numbers are stored in memory in the form of complement. In the complement of negative numbers, the first bit is 1 to represent the negative sign, so in the above loop, no matter how many loops are looped, it cannot be 0. The program infinite loop
It is a negative number before shifting, and it is still necessary to ensure that it is a negative number after shifting, so the highest bit will be set to 1 after shifting. If you keep doing right shifts, eventually the number will become
0xFFFFFFFF and fall into an infinite loop.
Idea 2: By bit manipulation and right shift operator
#include<stdio.h>
intmain()
{
int a=0;
int i=0,count=0;
printf("Please enter a positive integer:")
scanf("%d",&a);
for(i=0;i<32;i++)//
An integer occupies four bytes, and one byte is eight bits, so you need to loop 32
{
if((a>>i)&1==1)
{
count++;
}
}
printf("Number of 1's in binary in this integer:%d\n",count);
return 0;
}
Get the number of 1s in a binary number by controlling 1
#include<stdio.h>
int main()
{
int a=0;
int i=1,count=0,j=0;
printf("Please enter a positive integer:");
scanf("%d",&a );
while(i)
{
if (a&i)
{
count++;
}
i = i << 1;
}
printf("Number of 1's in the integer:%d\n",count);
return 0;
}
int main()
{
int a=0;
int i=1,count=0,j=0;
printf("Please enter a positive integer:");
scanf("%d",&a );
while(i)
{
if (a&i)
{
count++;
}
i = i << 1;
}
printf("Number of 1's in the integer:%d\n",count);
return 0;
}
Idea 3: By ANDing itself minus 1
#include<stdio.h>
intmain()
{
int a=0;
int i=0,count=0;
printf("Please enter a positive integer:")
scanf("%d",&a);
while(a)
{
a&=(a-1);
count++;
}
printf("Number of 1's in binary in this integer:%d\n",count);
return 0;
}
