Math.Round is being a bit annoying!

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So I've read the article in how to round numbers. And I've basically made a little sum for the console to print out. This is the sum:

C#

int rndx = (int)Math.Round((decimal)(53 / 32), 0, MidpointRounding.AwayFromZero);
Console.WriteLine(rndx.ToString());

Everytime I run this code, it returns as 'one'. While it should be two.

I've even checked my calculator which says that 53 / 32 = 1,65625. Which would be rounded up to 2 since it's over number 'x.5'.

Any answers?

Posted 23-May-13 9:19am

Deviant Sapphire

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2 solutions

Solution 1

Math.Round[^] returns decimal value, not integer! Follow the link!

See other methods:
Math.Floor()[^]
Math.Ceiling()[^]

Posted 23-May-13 9:31am

Maciej Los

Updated 23-May-13 9:33am

v2

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OriginalGriff 23-May-13 15:35pm

It's simpler than that: integer / integer = integer :laugh:

Maciej Los 23-May-13 15:36pm

I see it ;)

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OriginalGriff · Accepted Answer · 2013-05-23T09:32:00

I expect it to return one! :laugh:

Why? Simples:

53 and 32 are integers. so 53 / 32 is also an integer, value 1.

The double cast of an integer value will not restore lost fractional data...
Try:

C#

int rndx = (int)Math.Round(((decimal)53 / (decimal)32), 0, MidpointRounding.AwayFromZero);

Or

C#

int rndx = (int)Math.Round((53M / 32M), 0, MidpointRounding.AwayFromZero);

Which specifies decimal constants.

[edit]Fingers-typing-ahead-of-brain disease. "double" for "decimal" :doh: - OriginalGriff[/edit]