How to divide an integer by an integer and yield a decimal

Question

_{Tuesday, August 24, 2004 9:35 PM}

If we divide an integer by another integer, the result in C# is always an integer. How can we make the result to be 2 decimal? For example: 77 / 21 = 3.67 Thanks.

All replies (18)

_{Wednesday, August 25, 2004 7:16 AM}

Convert the integers to decimal: int i = 77; int j = 21; decimal d = (decimal)i / (decimal)j; Jim ThoughtWorks

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_{Wednesday, August 25, 2004 1:18 PM}

Thanks, that's what I found too. But the decimal needs to be formatted to 2 decimal. Otherwise it would show as a long bunch of decimals. Specifically how do we format a long decimal to "999.99" in C#? (We know how to do it in VB) Thanks.

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_{Wednesday, August 25, 2004 5:28 PM}

This is just saying for hell of saying, but you *could* always overload the \ operator. *grin*

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_{Wednesday, August 25, 2004 5:59 PM}

"you *could* always overload the \ operator" Overloads are attached to types, though. You can only overload operators on types that you control! Jim ThoughtWorks

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_{Wednesday, August 25, 2004 6:01 PM}

"how do we format a long decimal to "999.99"" Use the ToString() overload with a format specifier. Something like (off the top of my head): myDecimal.ToString("#.00"); Jim ThoughtWorks

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_{Friday, August 27, 2004 4:43 AM}

Also look at Math.Round, unfortunatly i think it only supports the double type in version 1.1 and decimal support is added in v2 Dave Legg

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_{Friday, August 27, 2004 5:26 PM}

I found it: myDecimal.ToString("N2"); Thanks.

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_{Sunday, August 29, 2004 4:13 AM}

To be honest, isn't the decimal type overkill if you are going for two places of precision when dividing an integer by an integer?

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_{Sunday, August 29, 2004 5:35 PM}

What we did here was for getting a decimal when one integer dividing another yields a decimal value such as 7/3=2.33, not a rounded integer. Thanks.

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_{Sunday, August 29, 2004 6:46 PM}

::Overloads are attached to types, though. You can only overload operators on types that you control! Jim, you're right, but you can control the output. you can return a different type. you can always overload the / for int,int to return a decimal type, and do the casting in the overloaded operator.

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_{Sunday, August 29, 2004 9:15 PM}

I realize the casting was done to retain the decimal amount, but I think a double would suffice.

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_{Monday, August 30, 2004 3:03 AM}

22/7... put that in a double. :)

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_{Monday, August 30, 2004 3:49 PM}

"you can always overload the / for int,int to return a decimal type" And how would you do that? Jim ThoughtWorks

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_{Monday, August 30, 2004 7:14 PM}

Kragie, they just want two decimal places.

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_{Tuesday, August 31, 2004 2:48 AM}

Jim, I didn't say it's going to be easy. Or that it was a good solution. I just said it was "possible". Just create an object that is essentially a decimal, and start creating a crapload of overloaded operators. Now. I'll admit that the original post was done in bad taste for humor, but still. It "is" possible. I'd be the first to slap someone that wanted to do it like that, but that wasn't the purpose of my initial post here.

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_{Tuesday, August 31, 2004 10:57 AM}

I'm interested in examples on overloads of operators. Please kindly provide some. Thanks.

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_{Thursday, September 21, 2017 6:00 PM}

I tried this, works for C#

int i = 77;
int j = 21;
double d = decimal.ToDouble(i) / j;
Console.WriteLine("{0:.00}",d);
Console.ReadLine();

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_{Tuesday, February 6, 2018 9:36 AM}

int i = 77;
int j = 21;
            
// convert int to decimal
decimal di = Convert.ToDecimal(i);
decimal dj = Convert.ToDecimal(j);

// compute 
decimal result = di / dj;

// display the result
Console.WriteLine("{0} / {1} = {2}", di, dj, result.ToString("n2"));