C / C++ / MFC

25-Sep-07 20:47

the value 0.00922307 should be rounded to 0.01.
Is there any method available in c++?

Nelek25-Sep-07 20:50

Nelek

25-Sep-07 20:50

Math.round ();

public static long round( double a )
Parameters
a
a double value.
Returns
the value of the argument rounded to the nearest long value.
Description
Returns the closest long to the argument.
If the argument is negative infinity or any value less than or equal to the value of Long.MIN_VALUE, the result is equal to the value of Long.MIN_VALUE.

If the argument is positive infinity or any value greater than or equal to the value of Long.MAX_VALUE, the result is equal to the value of Long.MAX_VALUE.

Greetings.

--------
M.D.V.

If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?

Help me to understand what I'm saying, and I'll explain it better to you

Wink | ;)

KASR125-Sep-07 20:54

25-Sep-07 20:54

I need to get result with two decimal places not long value.

megha_gharote25-Sep-07 20:58

megha_gharote

25-Sep-07 20:58

ROUND(nExpression, nDecimalPlaces)

Returns

Numeric

Arguments

nExpression

Specifies the numeric expression whose value is to be rounded.

nDecimalPlaces

Specifies the number of decimal places nExpression is rounded to.

If nDecimalPlaces is negative, ROUND( ) returns a whole number containing zeros equal in number to nDecimalPlaces to the left of the decimal point. For example, if nDecimalPlaces is –2, the first and second digits to the left of the decimal point in the value are 0.

This may work check once...

Megha

Cedric Moonen25-Sep-07 21:16

25-Sep-07 21:16

Which language is that ?
When you copy paste documentation that you found, it would be better to post the link instead.

Cédric Moonen
Software developer

Charting control [v1.2]

Cedric Moonen25-Sep-07 21:02

25-Sep-07 21:02

AFAIK, this is java libraries, not C++.

Cédric Moonen
Software developer

Charting control [v1.2]

Stephen Hewitt25-Sep-07 21:17

Stephen Hewitt

25-Sep-07 21:17

That looks like C#, not C++!

Steve

Nelek25-Sep-07 21:54

Nelek

25-Sep-07 21:54

I have taken this text from the VS 6.0 help, but without seeing from which library. It is java. Sorry for the confusion.

Cedric Moonen25-Sep-07 20:55

25-Sep-07 20:55

In general, rounding is needed when you want to display the number as a string. Is that the case for you ?
If yes, you can use sprintf and specify a precision:

char szBuff[25];<br />
sprintf(szBuff,"%.2f",yourDouble);

This tells sprintf to print 2 decimals after the floating point.

If you need to use it as a double (and not a string), you can multiply it by 100, convert it to an int and divided it back by 100.0:

double yourDouble = 0.00922307;<br />
int yourInt = yourDouble * 100;<br />
double roundedDouble = yourInt / 100.0;

(you must be carefull and divide by 100.0, not by 100, otherwise the you will end up with an integer and not a double).

Cédric Moonen
Software developer

Charting control [v1.2]

KASR125-Sep-07 21:03

25-Sep-07 21:03

but as per logic it returns 0.00 not 0.01 which required.

Nishad S25-Sep-07 21:09

Nishad S

25-Sep-07 21:09

Cedric Moonen wrote:
double yourDouble = 0.00922307;
int yourInt = yourDouble * 100;
double roundedDouble = yourInt / 100.0;

I think
double yourDouble = 0.00922307 + .005;
will be more right.

- NS -

Cedric Moonen25-Sep-07 21:13

Re: Rounding double value [modified]

25-Sep-07 21:13

Yes, you are perfectly right Big Grin | :-D

I forgot that the rounding is always rounded-down.

Cédric Moonen
Software developer

Charting control [v1.2]

Nelek25-Sep-07 21:51

Nelek

25-Sep-07 21:51

This is another way to do that.

I did it with VC++ 6.0 to be used in my project. The return value is a string to be shown in the GUI, but you can easily adapt it just for numbers.

CString CMyDoc::CutAndRoundStringToNDecimals (double dValue, const int nDec)
{	double dTemp = 0, dFract = 0, dInt = 0, dRes = 0;
	int nCount = 0;
	CString cLetter = "";

	// Operations to round up the nth decimal number if necessary
	dTemp = dValue*pow(10,nDec);
	dFract = modf (dTemp, &dInt);
	if (dFract >= 0.5)
		dInt++;
	dRes = dInt/pow(10,nDec);

	// Set the needed large of the string to be placed on the screen
	CString szResult;
	szResult.Format ("%s%f", szResult, dRes);
	int nLarge = szResult.GetLength ();
	for (int i = 0; i < nLarge; i++)
	{	nCount++;
		CString cLetter = szResult.GetAt (i);
		if (cLetter == ".")
		{	nCount += nDec;
			break;
		}
	}

	// Truncate the string to be placed in the screen
	LPTSTR pStr = szResult.GetBufferSetLength (nCount);
	szResult.ReleaseBuffer ();

	return szResult;
}

Hope it helps

I forgot to say that math.h has to be included to use pow () and modf()
-- modified at 11:41 Wednesday 26th September, 2007

beko25-Sep-07 23:57

beko

25-Sep-07 23:57

Hello,

i have come across with the int version of this function somewhere and changed to double in order to avoid precision problems for big and small values. You may use the int version if it fits your requirement.

double roundExt(double Num, double Seed)
{
// get rational factor
double fac = Num / Seed;

// get natural factor
fac = (fac < 0) ? (ceil(fac - 0.5)) : (floor(fac + 0.5));
// fac = (fac < 0) ? static_cast<int>(fac-0.5) : static_cast<int>(fac+0.5);

// return multiple
return fac * Seed;
}

roundExt(0.009,0.01); is the usage.

Hope this helps.

Bekir.

KASR126-Sep-07 1:15