Thread: Bitcount code

Hello,

I am trting to write a program that accepts numbers and count the number of bits in the given array of numbers. However my output is incorrect, I get the following

Enter nums:
10101010
The total count is 7

Code:

//Program that accepts a number and counts the bits.

#include <stdio.h>

int bitcnt(unsigned x);

int main (int argc, char *argv[]) {
    unsigned x;

    printf("Enter nums:\n");
    scanf("%d", &x);
//    bitcnt(x);

     printf("The total count is %d\n", bitcnt(x));

     return 0;
}


int bitcnt(unsigned x) {
    int b;

  for(b=0; b<x; x>>=1) {
    if(x&01) {
      b++;
     }
  }
//   printf("The total count is %d\n", b);
    return b;
}

All help is greatly appreciated.

Code:

int bitcnt(unsigned x) {
    int b;

  for(b=0; x; x>>=1) {
    if(x&1) {
      b++;
     }
  }
//   printf("The total count is %d\n", b);
    return b;
}

Ah ha ...

That makes sense now. Thanks alot for that.

Also is this the correct output?

Enter nums:
1001
The total count is 7

Thank You

Well, no - clearly there are 8 bits in a byte.

Counting like a C programmer, are you?

Well....Thanks for that

I dont quite understand why it is giving me that output and what i should do to modify it.

Thank You

Also i am new to programming, so my knowledge is a bit limited about C language.

Code:

    printf("Enter nums:\n");
    bitcnt += scanf("%d", &x);

// OR


    printf("Enter nums:\n");
    scanf("%d", &x);
    ++bitcnt;

1001 is 1111101001 in binary, so it has 7 1-bits in it and the answer is correct. Now you have to figure out if you've asked the correct question.

If you want to treat your input as a binary value, look into reading it as a string and then counting the number of 1s in that string.

If you want figure out how many bits in total, look at the sizeof() operator combined with the CHAR_BIT value defined in limits.h.