Learning loops: decimal to binary

[2hM9Nwbp]

I can't figure out why my code isn't working. I am trying to convert the users input from base 10 to binary. I went through it a few times and I just cant figure out why it isn't running properly. Any help would be much appreciated!

  /* This program reads from the user a positive integer
(in decimal representation), and prints its binary (base 2).*/
#include <iostream>
using namespace std;

int main() {

	int decimal, power, result, remain;

	cout << "Enter decimal number:" << endl;
	cin >> decimal;

	power = log(decimal)/ log(2);

	while (decimal > 0) {
		result = 2 ^ power;
		remain = decimal - result;
		if (remain >= 1) {
			cout << "1";
			decimal = decimal - result;
		}
		else {
			cout << "0";
		}
		power = power - 1;
	} 
	cout << endl;
	return 0;
}

[lastchance]

@capnrap,
I have to admit that I don't think this is a good way of doing the problem. A more common way would be to loop through looking at the end digit (decimal%2) and add character '0' or '1' at the start of a results string, followed by decimal /= 2 . Finally, print out the whole string.

However ... with your existing technique:
(1) result = 2 ^ power; doesn't do what you think it does (^ is a bitwise operator; you would need pow())
(2) Your test for looping is wrong - you should test on remaining power, not decimal, as the binary representation might end ...000 and you do need to write those zeroes.
(3) You should write a '1' if decimal/result is non-zero, not if remain >= 1 (which is usually true).
(4) log() and pow() are phenomenally expensive operations when you only need to be multiplying or dividing by 2 (which can, in fact, be accelerated by bitshifting).
(5) Consider using a separate function - you are tying down main() to just doing this single number conversion.

With the caveat that I REALLY DON'T THINK THIS IS A GOOD WAY OF DOING THE PROBLEM, minimal changes to your code are below. Note that it will only work for a number strictly greater than zero.

/* ***** This is NOT A GOOD WAY OF DOING THE PROBLEM **** */

#include <iostream>
#include <cmath>                            // you need this header for log() and pow()
using namespace std;

int main() {

	int decimal, power, result;

	cout << "Enter decimal number:" << endl;
	cin >> decimal;

	power = log(decimal)/ log(2);            // this is clunky ... and may well fail due to floating-point round-off

	while (power >= 0)                       // test on remaining positions, NOT residual number - you may have to write ...000
	{           
		result = pow(2,power);               // 2 ^ power does NOT do what you think it does
		if ( decimal / result !=0 ) {        // correct your test for a non-zero digit
			cout << "1";
			decimal = decimal - result;
		}
		else {
			cout << "0";
		}
		power = power - 1;
	} 
	cout << endl;
	return 0;
}

[jonnin]

its already in binary, FYI.
you can use bit operators to peel of the bits ... it displays in base 10, but its stored in binary.

[2hM9Nwbp]

Thank you so much.
I'll follow your advice and do the code the other way, but thank you for explaining why my code wasn't working.