C Programming - Bitwise Operators - Discussion
Discussion Forum : Bitwise Operators - Point Out Correct Statements (Q.No. 2)
2.
Which of the following statements are correct about the program?
#include<stdio.h>
int main()
{
unsigned int num;
int c=0;
scanf("%u", &num);
for(;num;num>>=1)
{
if(num & 1)
c++;
}
printf("%d", c);
return 0;
}
Answer: Option
Explanation:
If we give input 4, it will print 1.
Binary-4 == 00000000 00000100 ; Total number of bits = 1.
If we give input 3, it will print 2.
Binary-3 == 00000000 00000011 ; Total number of bits = 2.
If we give input 511, it will print 9.
Binary-511 == 00000001 11111111 ; Total number of bits = 9.
Discussion:
12 comments Page 2 of 2.
Pavan said:
5 years ago
T+=i&1 is same as T=T+(i&1) i.e,adding result (i&1) to initial value of T and again storing it in T.
Sibysagar said:
3 years ago
@Aami.
As we num>>=1 right shift the value of num, when it reaches binary value 0000000000000000 it gets out of for loop.
If we give num=4;
then the loop runs for 3 three times and the num value will be;
1st time entering loop num will be 4 (0000000000000100) after the right shift it will be(0000000000000010).
2nd time entering loop num will be 2 (0000000000000010) after the right shift it will be(0000000000000001).
3rd time entering loop num will be 1 (0000000000000001) after the right shift it will be (0000000000000000).
4th time entering loop num will be 0 (0000000000000000) so code will exit for a loop.
As we num>>=1 right shift the value of num, when it reaches binary value 0000000000000000 it gets out of for loop.
If we give num=4;
then the loop runs for 3 three times and the num value will be;
1st time entering loop num will be 4 (0000000000000100) after the right shift it will be(0000000000000010).
2nd time entering loop num will be 2 (0000000000000010) after the right shift it will be(0000000000000001).
3rd time entering loop num will be 1 (0000000000000001) after the right shift it will be (0000000000000000).
4th time entering loop num will be 0 (0000000000000000) so code will exit for a loop.
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