Cannot understand function logic

[whodgson]

the following is a small program from Brian Overland`s book "C++ In Plain English" It converts a number from decimal format to binary format as an example of bit shifting.
My problem is that i cannot follow the logic in the function.
here is the program:

#include<iostream>
#include <stdio.h>
 
void print_binary(short input_field);       //declare function print_binary
 
int  main (){
    printf("This program prints the binary equivalent of a short integer\n");
    printf("by AND_ing it with Ox8000 and assigning true (1) if the result\n");
    printf("is greater than 0 otherwise false (0)\n");
 
    short n ;                               // declare n as a short integer type
 
    do{                                     
        printf("\nEnter a short integer "); 
        printf("(0 to quit): ");            
        scanf("%hd",&n);                    
                print_binary(n);            // call function
  }  while (n);                             // continue to loop while n > 0
}    
void print_binary(short input_field) {      // function print_biary
    int i = 1,bit_set;                      //declare i (initialised to 1)
                                            // and bit_set
 
       while (i <=16) {                            //loop while i <= 16. get bit AND result
           bit_set=((0x8000 & input_field) > 0);   //this evaluates to 0 or 1.
                                                   // equiv of Ox8000 = 1000,0000,0000,0000
           printf("%hd",bit_set);                   //if 0 it prints bool 1 else 0
 
           input_field <<=  1;// move 1 bit to the left
           i++;
 
        }
}   

specifically AND (ing) OX8000 with an integer (<OX8000)
produces 0000 0000 0000 0000 so how can bit_set >0 ever be true (but even so how does it produce the binary equivalent of n)
I would appreciate however much excruciating detail is forthcoming.

[Rabbit]

I'll simplify the example.

n = 3;
while (i <= 4) {
   bit_set=((0x8 & input_field) > 0);
   input_field <<=  1;
   i++;
}

3 in binary is 0011.
0x8 in binary is 1000.

i = 1
input_field = 0011
bit_set = 1000 & 0011 = 0000 > 0 = 0
input_field <<= 1 = 0110

i = 2
input_field = 0110
bit_set = 1000 & 0110 = 0000 > 0 = 0
input_field <<= 1 = 1100

i = 3
input_field = 1100
bit_set = 1000 & 1100 = 1000 > 0 = 1
input_field <<= 1 = 1000

i = 4
input_field = 1000
bit_set = 1000 & 1000 = 1000 > 0 = 1

[whodgson]

thanks Rabbit
hmmm you say that (binary) 1000 & 0011 = 0011?
my book says 1000 & 0011 = 0000 or 0!
so does my calculator.
can we discuss further?

[Rabbit]

Oops, my fault, you're right, it is 0000. I fixed my example above.

[whodgson]

Yes...thanks Rabbit...like taking candy from kids