Program to count the number of bits set in c - c

Question

I have tried to count the number of bits set in an integer value in c.
But for some values it is showing the correct bit set count and for some values it is not.
PFB program code
int main()
{
int a=512,i=0,j=1,count=0,k=0;
for(i=0;i<31;i++)
{
if(k=a&j)
{
count++;
j=j<<1;
}
}
printf("the total bit set countis %d",count);
}
The output of set bit value count of 512 is showing as zero and if the value used is 511 count is showing as 9.
Please help me to correct the program.

Answer 1

Stanford University has a page of different ways to implement common bit-twiddling operations. They list 5 different algorithms to count the bits set, all with C examples.
https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive
Their simplest implementation:
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; v >>= 1)
{
c += v & 1;
}

Answer 2

If you're using gcc/clang compiler, you can use the builtin function __builtin_popcount
unsigned int user_input = 100
int count = __builtin_popcount(n); // count == 3
When I'm not looking for cross-platform I'll use this function since its highly optimised.

Answer 3

Generally you would count bits in an unsigned integer. The reason being that you're usually checking for bits set in a register or a mask, for example. Signed integers are represented using twos-compliment and I can't think why you'd want to count set bits in a signed integer (would be interested why if you definitely do want this).
Note in C that right or left shifting a signed integer is implementation defined behaviour if the number is negative. From C standard sectn 6.5.7:
... The result of E1 << E2 is E1 left-shifted E2 bit positions; ... If E1
has a signed type and nonnegative value, and E1 << E2 is representable
in the result type, then that is the resulting value; otherwise, the
behavior is undefined.
The result of E1 >> E2 is E1 right-shifted E2
bit positions. ... If E1 has a signed type and a negative value, the
resulting value is implementation-defined ...
If you want to count 1's in an arbitrary sized unsigned integer you could use this example:
#include <stdio.h>
int main(void) {
unsigned int value = 1234;
unsigned int ones = 0;
while(value > 0) {
ones += value & 0x1;
value >>= 1;
}
printf("#Ones = %u", ones);
}
Using this example value could be unsigned char, unsigned long, whatever unsigned integer type...
Note: Do not shift signed values or floats/doubles.

Answer 4

You can use the division / and the modulo % operator to check the bits that are set in an integer.
int main()
{
int a = 512, count = 0;
while(a != 0)
{
if(a % 2 == 1)
{
count++;
}
a /= 2;
}
printf("The total bit set is %d", count);
}

Answer 5

You have a couple of mistakes:
for(i=0;i<32;i++) // <<< this should be 32, not 31
{
if(k=a&j)
{
count++;
}
j=j<<1; // <<< this needs to be outside the if block
}
Note that instead of using a hard-coded value of 32 for the no of bits in an int, it would be better to do it like this:
for(i=0;i<sizeof(int)*CHAR_BIT;i++)
This way the code will still work if the size of an int is e.g. 16 bits or 64 bits.

Answer 6

Although this is not C strictly speaking, you can use inline assembly to call the POPCNT x86 operation:
// GCC syntax
unsigned a = 1234;
unsigned int count;
__asm__(
" POPCNT %0, %1\n"
:"=r" (count)
:"r" (a)
);
return count;
According to this benchmark, calling __builtin_popcount as in idok's answer is just as fast as the above code and they both are much faster than any other C implementation. You can also check the linked repo for other solutions as well.

Answer 7

You are checking the value of a&j , and if a&j is 0, then you do nothing else but try again.
Your j-bitshift needs to be outside the if-then.

Answer 8

#include<stdio.h>
#include<conio.h>
int rem, binary = 0;
unsigned int
countSetBits (unsigned int n){
unsigned int count = 0;
while (n){
count += n & 1;
n >>= 1;
}
printf ("\n\t Number of 1's in the binary number is : %d",count);
}
int dec_bin (int n){
int i=1;
while (n != 0){
rem = n % 2;
n = n / 2;
binary = binary + (rem * i);
i = i * 10;
}
printf("\n\t The converted Binary Equivalent is : %d",binary);
}
int main(){
int i = 0;
printf ("\n\t Enter the Decimal Nummber: ");
scanf ("%d", &i);
int n= i;
dec_bin(n);
countSetBits (i);
return 0;
}