Thread: understanding output

try this
char ch='%';

printf("%d %c = %d", n, ch, (n%2));

I was able to produce this code:

#include <stdio.h>

void puzzle(int n) {

if(n) {
printf("n = %d\n",n);
printf("%d %% 2 = %d\n", n, n%2);
/* To display "%" on the screen, use the escape seq %% */
puzzle(n/2);


/* What is up with the following lines??? */
putchar('0' + n%2);
printf("\nn now = %d\n", n%2 );
/* Maybe with some explaination, I could help more, like what
is this program supposed to do exactly?? */

}
}

main()
{
int i;
printf("input value: ");
scanf("%d",&i);

puzzle(i);
printf("\nValue after function call = %d\n\n",i);

return 0;
}


Maybe post something about what you are trying to accomplish

Actually you would do this:
printf("%d %% 2 = %d\n",n,(n%2));
% is the escape for the % sign.

Basically the code prints out the binary equivalent to the number input by user. But I am failing to see how it prints a 1, exactly. I think when with the expression

putchar('0' + n%2);

the code prints out a 0 if n%2 = 0 (thus 0 + 0 = 0)
or prints out a 1 if n%2 = 1 (thus 0 + 1 = 0)
Is this right??

What I don't get is what is printed out (what 'n' is before and after) this putchar statement and right before it recursively calls the function again.

This recursive stuff gets me a little uneasy, and I know from past exams the instructor will test heavily on this. So I want to get a good idea of this simple problem.

Thanks for any ideas, help, suggestions....