Question

I’m doing simple divisions in c#, and I am a bit puzzled by its intricacies. Here’s some code, and in the comments, the result. (btw, I only compile with 1 line not commented, if you say that I have 5 declarations of the same variable)

double result = 2 / 3; //gives 0
double result = Convert.ToDouble(2) / Convert.ToDouble(3); // is good
double result = double.Parse(2) / double.Parse(3); // gives me errors
double result = double.Parse(2 / 3); // gives me errors
double result = Convert.ToDouble(2 / 3); // gives 0

MessageBox.Show(result.ToString());

so if you have a bunch of integers you wanna mess with, you have to convert each one to a double. pretty tedious…

Answer 1

Integer division discards the remainder. You don’t need to use Convert.ToDouble or double.Parse, you can simply write:

double result = 2.0 / 3;

or

double result = (double)2 / 3;

If either one of the operands is a floating-point value then you get floating-point arithmetic instead of integer arithmetic.

To explain each one:

// 2 / 3 = 0 with a remainder of 2; remainder is discarded
double result = 2 / 3;

// 2.0 / 3.0 = 0.6667, as expected
double result = Convert.ToDouble(2) / Convert.ToDouble(3);

// double.Parse takes a string parameter, so this gives a syntax error
double result = double.Parse(2) / double.Parse(3);

// As above, double.Parse takes a string, so syntax error
// This also does integer arithmetic inside the parentheses, so you would still
// have zero as the result anyway.
double result = double.Parse(2 / 3); // gives me errors

// Here, the integer division 2 / 3 is calculated to be 0 (throw away the
// remainder), and *then* converted to a double, so you still get zero.
double result = Convert.ToDouble(2 / 3);

Hope that helps explain it.