Help counting number of bits set in an integer

I have to write a program that counts the number of bits set in an integer. For example, the number 5 (decimal), which is 0000000000000101 (binary), has two bits set. I know how to count the number of bits in a char, but I don't know how to count the extra 8 bits in an integer. How would i navigate to the other byte and count the eight bits?

Here is how i'd count the number of bits in a char...

Code:

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#include <stdio.h>

void main()
{
        char ch = 5;  /*assign it a random number*/
        int counter = 0;
        int i;
       
        for (i = 0x80; i > 0; i = (i >> 1))  {
            if ((ch & i) != 0)
                ++counter;
        }


}

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Thanks for any help with this...

Code:

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while(i)
{
  counter += i & 0x1;
  i >> 1;
}

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> void main()
main returns an int

> How would i navigate to the other byte and count the eight bits?
Other 8 ?
The size of an int varies from machine to machine, as do all the other types.

If you include limits.h, then the number of bits in an int is
CHAR_BIT * sizeof(int)

> for (i = 0x80;
Well the easy thing to do is start with a different constant which starts with the most significant bit of the int.

Hint
unsigned int allOnes = ~0;
Using that and some bit operations, you should be able to find the value of the most significant bit in an int without explicitly saying it in your code.

Code:

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i >> 1;

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Code:

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i >>= 1;

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little typo vart

Quote:

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little typo vart

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Thanks

Right shift on signed numbers is implementation-defined behaviour.