Overloaded function Modulus

[rnima]

I need big help with figuring out Int operator%(const Int&); i havent started on division just yet because it looks like its going to be more complicated
This is my Class

#ifndef INT_H
#define INT_H

#include <iostream>
#include <cmath>

using std::ostream;
using std::istream;
using std::string;

class Int {
private:
    short* number;
    int     size;
    void    computeSize();
    bool negative;

public:
    Int(long = 0);
    Int(const char*);
    Int(const Int&);
    ~Int();
    
    Int operator/(const Int&);
    Int operator%(const Int&);

Int Int::operator%(const Int& right)
{
	cout << "operator%(const Int&)" << endl;
        // i need help here not sure at all how to do this please guide me through 
}

Int::Int(long value)
{
	cout << "Int(long) called" << endl;
	this->negative = false;
	size = 0;
	number = new short[30];

	for (int i = 0; i < 30; i++)
		number[i] = 0;

	for (int i = 29; value != 0 && i >= 0; i--) {

		number[i] = value % 10;
		value /= 10;
	}
}
Int::Int(const char* string)
{
	cout << "Int(const char*) called" << endl;
	number = new short[30];
	size = 0;
	this->negative = false;
	for (int i = 0; i < 30; i++)
		number[i] = 0;

	int length = strnlen(string, 30);
	for (int j = 30 - length, k = 0; j <= 29; j++, k++) {

		if (isdigit(string[k]))
			number[j] = string[k] - '0';
	}
}
Int::Int(const Int& copy)
{
	cout << "Int(const Int&) called" << endl;
	number = new short[30];
	size = 0;
	this->negative = false;
	
	for (int i = 29; i >= 0; i--)
		number[i] = copy.number[i];

}

Please please help me, I'm quite new to overloaded operators

[helios]

Division and remainder are actually extremely similar.

A simple algorithm:

remainder = 0
while dividend is not zero
    remainder = remainder * 2 + dividend % 2
    dividend /= 2
    if remainder >= divisor
        remainder -= divisor
return remainder

[rnima]

So i can't follow through with using %, i get an error saying "No operator matches these operands" How can i apply this logic with my code ? if i am taking "right " as a parameter, then right would be my divided ? or would number be my dividend ?

[jonnin]

a few observations:
division is usually rather complicated, but multiply usually is not. Granted, both are repeated addition (subtraction is just addition of a negative number) if you want to go brute force at it. If you can rewire division into a multiply, it may be easier to implement.

if you had division, the remainder can be had without loops, just a few short steps:
say you had ... 1023 % 13 for example.
1023/13 is 78 integer division.
13*78 is 1014.
1023-1014 is 9.

[helios]

So i can't follow through with using %

Of course not. But x % 2 is simply the least significant bit of x, which should be trivial to obtain.
x / 2 is simply all the bits of x shifted down towards the least significant end by 1 digit.
(Obviously x * 2 is all the bits shifted up by 1.)

if i am taking "right " as a parameter, then right would be my divided ? or would number be my dividend ?

*this would be the dividend (the left-hand side of the operation) and the parameter would be the divisor (the other side).

[againtry]

For 'ordinary' numbers division or modulo arithmetic can be carried out by repeated subtraction. So if you can do subtraction (and addition) you can do the rest.

#include <iostream>

// dividend/divisor = quotient (+ remainder)

int main()
{
    int dividend{1023};
    int divisor{13};
    int quotient{0};
    int remainder{0};
    
    while( dividend >= divisor)
    {
        quotient ++;
        dividend -= divisor;
    }
    
    remainder = dividend;
    std::cout << quotient << ' ' << remainder << '\n';
    
    return 0;
}