Code for counting no. of 0s and 1s in binary representation of a letter

Basically, I have to write a code such that:

1. a letter is taken as an input.
2. no. of 0s and 1s in the binary representation of it’s ascii value are counted
3. report (no. of 0s)*(no. of 1s)

Please help me find mistakes in my code.

#include <stdio.h>
int main(){
    
    int x,y,count0,count1,a,r;
        scanf("%c",&x);
        a = (int)x;

        count0 = 0;
        count1 = 0;
        while(a != 0){
            r=a%2;
            if(r == 0) 
                count0 = count0 + 1;
            else if(r == 1) 
                count1 = count1 + 1;
            a = a/2;
        }
        y = count1 * count0;
        printf("%d\n",count0);
        printf("%d\n",count1);
        printf("%d",y);
        return 0;
}

Clearly this is wrong as for the input ‘A’, it should print 5,2,10 but it is giving out 16,7,112.

>Solution :

Because you declared x as int, scanf("%c",&x); produces:

test.c:5:17: warning: format ‘%c’ expects argument of type ‘char *’, but argument 2 has type ‘int *’ [-Wformat=]

Also default char is signed, which may produce surprising numbers of ones and zeroes if you encounter one of the negative values.

Change the type of x to be unsigned char:

unsigned char x;
int y,count0,count1,a,r;

then the code should do what you wanted.

If you don’t have warnings enabled in your compiler, enable them now. They are usually helpful in detecting problems like this.