How Does the Compound Modulo Work in C++?

Question

I have been trying to create simple program which divides an input number into peso bills. The output I need is

Enter Amount: 7350
P1000: 7
P500: 0
P200: 1
P100: 1
P50: 1
P20:0
P10:0
P5:0
P1:0

and this was my initial code:

#include <iostream>
using namespace std;

int main()
{
  int const P1000(1000);
  int const P500(500);
  int const P200(200);
  int const P100(100);
  int const P50(50);
  int const P20(20);
  int const P10(10);
  int const P1(1);
  
  int input;
  //input number
  cout<<"input number in pesos: ";
  cin>>input;
  cout<<"P1000 = "<<input/P1000<<endl;
  cout<<"P500 = "<<(input%1000)/P500<<endl;
  cout<<"P200 = "<<(input%500)/P200<<endl;
  cout<<"P100 = "<<(input%200)/P100<<endl;
  cout<<"P50 = "<<(input%100)/P50<<endl;  
  cout<<"P20 = "<<(input%50)/P20<<endl;
  cout<<"P10 = "<<(input%20)/P10<<endl;
  cout<<"P1= "<<(input%10)/P1<<endl;
  return 0;
}

but the output I get is

input number in pesos: 7350
P1000 = 7
P500 = 0
P200 = 1
P100 = 1
P50 = 1
P20 = 0
P10 = 1
P1= 0

so after a bit of tinkering I used this block of code which worked wonders!

#include <iostream>
using namespace std;

int main()
{
  int const P1000(1000);
  int const P500(500);
  int const P200(200);
  int const P100(100);
  int const P50(50);
  int const P20(20);
  int const P10(10);
  int const P1(1);
  
  int input;
  //input number
  cout<<"input number in pesos: ";
  cin>>input;
  cout<<"P1000 = "<<input/P1000<<endl;
  cout<<"P500 = "<<(input%=1000)/P500<<endl;
  cout<<"P200 = "<<(input%=500)/P200<<endl;
  cout<<"P100 = "<<(input%=200)/P100<<endl;
  cout<<"P50 = "<<(input%=100)/P50<<endl;  
  cout<<"P20 = "<<(input%=50)/P20<<endl;
  cout<<"P10 = "<<(input%=20)/P10<<endl;
  cout<<"P1= "<<(input%=10)/P1<<endl;
  return 0;
}

So my question is, how why did using the compound modulo operator work? How is it different from the regular modulo operator? I don't think the math is the problem but the way the code is handled. This is my first few weeks of learning C++ (and programming in general) and it would be nice to clear up some of my confusion. Thank you in advance.

LIke with any other compound assignment operator, it assigns the result of the operation to the left side in the equation. — πάντα ῥεῖ, Oct 27 '20 at 06:33

score 2 · Accepted Answer · answered Oct 27 '20 at 08:00

Here is a small program to illustrate the difference:

#include <iostream>
int main()
{
    int i = 256;

    std::cout << "i: " << i << ", i % 100: " << (i % 100) << std::endl;
    std::cout << "i: " << i << ", i % 12: " << (i % 12) << std::endl;
    std::cout << "i: " << i << ", i % 3: " << (i % 3) << std::endl;
    
    std::cout << std::endl;

    std::cout << "i: " << i << ", i %= 100: " << (i %= 100) << std::endl;
    std::cout << "i: " << i << ", i %= 12: " << (i %= 12) << std::endl;
    std::cout << "i: " << i << ", i %= 3: " << (i %= 3) << std::endl;
}

This outputs (fixed spaces to be nicer):

i: 256, i % 100: 56
i: 256, i % 12:   4
i: 256, i % 3:    1

i: 256, i %= 100: 56
i: 56,  i %= 12:   8
i: 8,   i %= 3:    2

As you can see in the first part, the regular modulo operator leaves i at the original value. This means that we get 256 modulo 100, 12 and 3.

However, in the second part the compound modulo operator keeps changing i, so that the next modulo only operates on the remainder.

How Does the Compound Modulo Work in C++?

1 Answers1