C# double not working as expected

Question

I understand that a double is a decimal. In the following program the output is 1 even though I thought it would be 1.05 repeating.

static void Main (string[] args)
{
double d = 19 / 18;
Console.WriteLine(d);
Console.ReadKey();
}

Am I misunderstanding double?

You are misunderstanding integer math: `Integer-19` divided by `Integer-18` results in a Integer: `1`. Try `19.0 / 18.0` — abelenky, Mar 26 '15 at 20:11
"I understand that a double is a decimal" - not really. It's a binary floating point number, although that's not the problem you've got. Your actual problem is you're performing integer division, then converting the result (an integer) into a double... — Jon Skeet, Mar 26 '15 at 20:12
[C# Reference Division Operator](https://msdn.microsoft.com/en-us/library/aa691373(v=vs.71).aspx) — Steve, Mar 26 '15 at 20:12
`double d = 19 / 18;` -> `double d = (double)(19 / 18);` -> `double d = (double)(1);` -> `double d = 1.0d;` — Zéychin, Mar 26 '15 at 20:20

score 3 · Accepted Answer · answered Mar 26 '15 at 20:13

3

You are misunderstanding integer math.

Integer-19 / Integer-18 results in an Integer with value 1.

(that you assign the value to a double is not relevant. The calculation results in an Integer).

To fix it, use:

double d = 19.0 / 18.0;

answered Mar 26 '15 at 20:13

abelenky

or double d = 19d / 18d; – Xela Mar 26 '15 at 20:25
@Xela or `double d = 19.0 / 18` or `double d = 19 / 18.0` or `double d = 19d / 18` or `double d = 19 / 18d` :) – EZI Mar 26 '15 at 20:35
or overkill double d = Convert.ToDouble(19) / Convert.ToDouble(18); lol – Xela Mar 26 '15 at 20:38

1 Answers1