Hi all,
I need to round a decimal value with a particular rule.
For example:

decimal a = 1.49m -> 1m
decimal a = 1.5m -> 1m
decimal a = 1.51m -> 2m

I've tried Math.Round but with the value 1.5 it returns me 2, which it is
not correct for my business rule.

How can I solve this problem?

Thanks in advance.


--
Luigi

How about something cheeky like:

decimal a = Math.Ceiling(x-0.5M);

Obviously you'd need to think about -ves etc...

Marc

On Jul 1, 3:23*pm, Luigi <ciupazNoSpamGra...@inwind.it> wrote:
> I need to round a decimal value with a particular rule.
> For example:
>
> decimal a = 1.49m -> 1m
> decimal a = 1.5m -> 1m
> decimal a = 1.51m -> 2m
>
> I've tried Math.Round but with the value 1.5 it returns me 2, which it is
> not correct for my business rule.
>
> How can I solve this problem?

I'd write a custom rounding method. It's reasonable simple if you have
a very specific requirement. In this case, something like:

public decimal Round (decimal value)
{
decimal floor = Math.Floor(value);
decimal ceiling = Math.Floor(value);
decimal midpoint = (floor+ceiling)/2;
return value <= midpoint ? floor : ceiling;
}

It's possible that there will be some weird problems around the very
largest numbers that decimals can store, but if your application
doesn't use those (and it's unlikely to) then you should be okay.

This is completely untested though - I strongly recommend writing unit
tests for it!

Jon

On Jul 1, 10:23*am, Luigi <ciupazNoSpamGra...@inwind.it> wrote:
> Hi all,
> I need to round a decimal value with a particular rule.
> For example:
>
> decimal a = 1.49m -> 1m
> decimal a = 1.5m -> 1m
> decimal a = 1.51m -> 2m
>
> I've tried Math.Round but with the value 1.5 it returns me 2, which it is
> not correct for my business rule.
>
> How can I solve this problem?
>
> Thanks in advance.
>
> --
> Luigi

Try multiplying by 10*(number of decimals needed), used math.floor or
math.ceiling, then divide by same number 10*(number of decimals
needed)

Assuming you mean 10 raised-to-the-power-of (number of decimals), the OP
appears to want zero decimals. Taking the cited example 1.51, this would
give 1M, not 2M as desired.

Marc

rossum <(E-Mail Removed)> wrote:

<snip>

> > decimal ceiling = Math.Floor(value);
> Did you mean "decimal ceiling = Math.Ceiling(value);"? rossum

Yup. That's what I get for not testing it...

--
Jon Skeet - <(E-Mail Removed)>
Web site: http://www.pobox.com/~skeet
Blog: http://www.msmvps.com/jon_skeet
C# in Depth: http://csharpindepth.com

Thank you all very much.
I'll try this one method:

decimal x = 1.51m;

decimal a = Math.Ceiling(x - 0.5M);

Luigi

Note that this is called the banker rounding: if, for x a positive integer,
x + .5 is always rounded to (x+1), the banker ends up loosing money. So,
1.5 is rounded to 2, that is, upward, the banker lost is 0.5; while 2.5 is
*also* rounded to 2, that is, downward, and the banker get a profit of 0.5
in that case.

Statistically, the banker should be not loosing, neither winning (assuming
the integer part has no bias toward odd or toward even numbers) with that
way to round. You probably already know it, but just in case someone reading
this thread was not.

Vanderghast, Access MVP

"Luigi" <(E-Mail Removed)> wrote in message
news:F676C879-12A5-4B4E-B8D5-(E-Mail Removed)...
> Hi all,
> I need to round a decimal value with a particular rule.
> For example:
>
> decimal a = 1.49m -> 1m
> decimal a = 1.5m -> 1m
> decimal a = 1.51m -> 2m
>
> I've tried Math.Round but with the value 1.5 it returns me 2, which it is
> not correct for my business rule.
>
> How can I solve this problem?
>
> Thanks in advance.
>
>
> --
> Luigi
>

> Note that this is called the banker rounding

I'm fairly certain that the original question as posed doesn't relate
to banker's rounding; it is just .5 always rounding down.

Marc

Yes, sorry, I was not clear. The OP want a rounding down at 0.5, but GOT a
banker rounding, with Math.Round(x, 0).

Vanderghast, Access MVP


"Marc Gravell" <(E-Mail Removed)> wrote in message
news:c1e60763-827e-44a7-9cfc-(E-Mail Removed)...
>> Note that this is called the banker rounding
>
> I'm fairly certain that the original question as posed doesn't relate
> to banker's rounding; it is just .5 always rounding down.
>
> Marc