Dividing integers and remainders

I would really like to se an integer (var1) containing a fractional number :D

Quite clearly you are a lunatic if you think an integer can
contain .0333333333333333! :-)

Adrian

Using the following program, I get:

2 3
2 4

#include <iostream>
using namespace std;

void Function1(int var1, int var2) {
int quotient = var2 / var1;
int remainder_times_100 = (100 * (var2 - var1 * quotient)) / var1;
cout << quotient << ' ' << remainder_times_100 << '\n';
}

void Function2(int var1, int var2) {
div_t Div = div(var2, var1);
cout << Div.quot << ' ' << Div.rem << '\n';
}

int main(int argc, char *argv[], char *envp[]) {
Function1(120, 244);
Function2(120, 244);
return 0;
}

As far as I know, floating-point is unreliable for things like this.

Sam:
Your Function2 returns the remainder of the division, which wasn't the question as I understood it. My understanding was that "hoagers34" wanted the first two digits of the fractional part of the division, ie., the "03" of ".03333333".

Once again I have to ask: Why isn't the original questioner taking part in this discussion?

I would like to know why hoagers34 wishes to perform this operation.

Is it so that further calculations can be made using the adjusted figure, say for currency calculations, or is it purely to make the value display properly in a text box?

If it's the first case, I would recommend working in pennies rather pounds/dollars/etc and then you only need to work with whole numbers.

If it's for display reasons, use whatever formatting operators your output method has for limiting the displayed value to 2 decimal places.

Either way, it's quicker and simpler than trying to manipulate a floating point variable.

Many of the threads that grow beyond reason are those in which the creator of the thread does not participate after creating it. Without the participation of hoagers34 I am not interested in speculation and will simply unsubscribe from this thread.