Rounding a negative value is zero?

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Hello everyone!

I am currently trying round figure out how much a value out of 20 is. For instance,

-5 out of 20 would be -1.
5 out of 20 would be 1.

25 out of 20 would be 2.
-25 out of 20 would be -2

However, when I do that:

C#

int newValue = (int)Number / 20

-5 returns as 0. Which is normal, however, I need the negative value -1. Does anyone have any ideas? This is driving me insane, and doing this manually:

C#

if(Number.StartsWith "-")

Would not make any logic for me, and not for my program either. Please help!

Posted 3-May-13 10:43am

Yvar Birx

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lewax00 3-May-13 17:06pm

I don't really get what you're trying to do here, but integer division always truncates. So: -5/20 = -0.25 which will truncate to -0, which is the same as 0. As far as checking if a value is negative though, just check if it's less than 0.

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André Kraak · Accepted Answer · 2013-05-03T11:04:00

Solution 1

For positive numbers use Math.Ceiling Method (Double)[^] and for negative numbers use Math.Floor Method (Double)[^].
So:

C#

const double newValue = Number / 20.0;
double outOfTwenty = 0.0;

if (newValue >= 0.0)
    outOfTwenty = Math.Ceiling(NewValue);
else
    outOfTwenty = Math.Floor(NewValue);

Posted 3-May-13 11:04am

André Kraak

Comments

Sergey Alexandrovich Kryukov 3-May-13 17:07pm

Right, a 5. You could also have mentioned Math.Round, for comparison.
—SA

Yvar Birx 3-May-13 17:18pm

-16 returns to 0?

-16 / 20 = -0.8?

André Kraak 3-May-13 17:22pm

Math.Floor(-0.8) should be -1.0

Yvar Birx 3-May-13 17:23pm

if (Number > 0)
{
Number = Math.Ceiling(Number );
}
else
{
Number = Math.Floor(Number );
}

Yvar Birx 3-May-13 17:50pm

:S

Ian A Davidson 3-May-13 17:56pm

You need to work with double, not int. Conversion to int always loses the decimal place. Thus:
int x = -16 / 20; // x will be 0
double y = (double)-16 / (double)20; // y will be 0.8.

Note that at least one of the operands in the expression also need to be double (in this case cast to double) - it doesn't hurt to ensure both are, since the compiler will do the cast anyway. Otherwise integer arithmetic will be used and then simply assigned to the double. I.e.
double z = -16 / 20; // z will also be 0

Specifying the decimals part also causes it to interpret the value as a double, hence the answer is right.

double y = -16 / 20.0; // y will be 0.8.

By the way, I don't understand your question either, because 5 / 20 will also give the answer of zero, not one.

Regards,
Ian.

Yvar Birx 3-May-13 18:31pm

Okay, so let's suppose everything is positive.
Why does (15) return as 0?

int Value = (int)Math.Ceiling((double)(Math.Abs(15) / 20));

Yvar Birx 3-May-13 18:31pm

15 / 20 = 0.75

Yvar Birx 3-May-13 18:32pm

Sorry for the abs, I had a negative number inside it like a few seconds ago.

Yvar Birx 3-May-13 18:34pm

And then rounding 0.75 up would equal to one? Even if it's just more than 0.5?

Ian A Davidson 3-May-13 18:40pm

Because, as I said, it's doing integer arithmetic so loses everything after the decimal point.

15 / 20 = 0
15 / 20.0 = 0.75

Ian.

Yvar Birx 3-May-13 18:51pm

Much obliged!

Ian A Davidson 3-May-13 18:53pm

You would need to do this instead:

int value = (int)Math.Ceiling(Math.Abs(15) / 20.0);

I've left your call to Abs so you can see the difference.

Rounding a negative value is zero?

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Rounding a negative value is zero?

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