Программа на языке Си для преобразования десятичной дроби в двоичную с помощью побитового оператора and, shift

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this code is an example from a book that the problem require to change decimal to binary number using bitwise AND oeprator and shift operator. i cannot understand the code although had tried to understand this code using debug compiler. suppose for a and b, user input is 10 and 8

#include <stdio.h>
#include <stdlib.h>


int count_bits (unsigned x)
{

    int bits=0;
    while(x){
        if (x&1U)bits++;
            x>>=1;


    } return bits;
}

int int_bits(void)
{
    return count_bits(~0U);
}

void print_bits(unsigned x){
    int i;
    for(i=int_bits(x)-1;i>=0;i--)
        putchar(((x>>i)&1U)?'1':'0');
}
int main(void)
{
 unsigned a,b; /*suppose user input a=10 b=8*/
 printf("enter two positive integer value=\n");
 printf("a=  "); scanf("%u",&a);
 printf("b:  "); scanf("%u",&b);

 printf("\na   =");  print_bits(a);
 printf("\na   =");  print_bits(b);
    return 0;
}

Что я уже пробовал:

in int_bits function what actually (~0U) does? <pre>
 what is the value of x in count_bits(unsigned x) function? and what is being compare in (x & 1U) and what is the relation of count_bits(~0U) and user input?
in print_bits function in putchar (((x>>i)&1u)?'1';'0'); what is the value of x and i? 
i got i=32 as bits from count_bits

что на самом деле делает эта программа для получения двоичного числа?

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