C / C++ / MFC

29-Sep-06 18:48

Hello,

I have a floating point variable:

float fMyFloat;

How do I round it too a desired number of decimal places?

Thanks,
BP

-- modified at 14:38 Saturday 30th September, 2006

ShilpiP29-Sep-06 19:14

ShilpiP

29-Sep-06 19:14

u use %.2f format specifier in place of %f
float fMyFloat;
printf("%.2f",fMyFloat);
Smile | :)

amaneet29-Sep-06 21:42

amaneet

29-Sep-06 21:42

Sir,
I want to do the sane thing in vc++ I mean i don't want to use printf .I mean i have to round a float variable .Can i know a function that do this functionality.
Thanks

ShilpiP29-Sep-06 23:14

ShilpiP

29-Sep-06 23:14

ok u may use floor function .....i m sending u one link
http://forums.dreamincode.net/showtopic8568.htm[^]
may be it will help u
Smile | :)

Stephen Hewitt30-Sep-06 2:27

Stephen Hewitt

30-Sep-06 2:27

First you have to decide what kind of rounding you’re after. There are four common ways rounding can be performed:
- Towards +ve infinity.
- Towards -ve infinity.
- Towards 0.
- Away from 0.

Towards +ve infinity:
In this mode we choose the closest number that’s higher. For example, if rounding to the nearest integer:
1.6 -> 2
-1.6 -> -1

Towards -ve infinity:
In this mode we choose the closest number that’s lower. For example, if rounding to the nearest integer:
1.6 -> 1
-1.6 -> -2

Towards 0:
In this mode we chose the closet number in the direction of zero. For example, if rounding to the nearest integer:
1.6 -> 1
-1.6 -> -1

Towards 0:
In this mode we chose the closet number in the direction away from zero. For example, if rounding to the nearest integer:
1.6 -> 2
-1.6 -> -2

For simplicity I’ll only consider the “Towards +ve infinity” mode. If you want to round a double or float to the closest integer code like this will do it:
int n = double_val+0.5;

To a double to the nearest tenth:
double n = floor(double_val*10.0+0.5)/10.0;

To a double to the nearest hundredth:
double n = floor(double_val*100.0+0.5)/100.0;

Steve

Hamid_RT2-Oct-06 9:02

Hamid_RT

2-Oct-06 9:02

See here[^]

WhiteSky

Eytukan30-Sep-06 7:32

Eytukan

30-Sep-06 7:32

use floor and ceil

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BlitzPackage30-Sep-06 10:05

30-Sep-06 10:05

First, thank you all for your answers, they are proving very helpful.

Now, however, I've encountered a different issue:

I get the right results from the floor function. However, when I do the final division (I'm getting this information from stepping through the code in the VC++ .NET Debugger):

//m_fAmount is a float
m_fAmount = 30679.000
iDivisor = 1000.0000

m_fAmount / iDivisor yields: 30.679001

I am really puzzled as I'd like to think that the compiler can do simple arithmatic Smile | :)

However, in all seriousness, what's causing this, I'm really puzzled.

Any and all help is greatly appreicated.

Thanks,
BP

markkuk30-Sep-06 12:52

markkuk

30-Sep-06 12:52

It's caused by round-off errors in decimal->binary->decimal conversion.

BlitzPackage30-Sep-06 13:36

30-Sep-06 13:36

Do you know where I can find some information on that error and how to compensate?

Thanks,
BP

Stephen Hewitt30-Sep-06 16:04

Stephen Hewitt

30-Sep-06 16:04

Just say we're working in decimal with 4 places after the decimal point. Using this scheme we can represent the following numbers exactly:
1.0000
3.0000

But it we try to divide them:
1.0000/3.0000 = 0.3333

This answer is wrong already, the correct answer us 0.333333.... (the 3 recurs forever).

So now consider this sum:
(1.0000/3.0000)*3.0000 = 0.9999

This sort of things happens in all floating or fixed point numbering systems. The computer uses base 2 and not base 10 however.

Try using double instead of float; there is seldom a good reason to prefer a float over a double

Steve

BlitzPackage30-Sep-06 19:11

Return an Object By Value

30-Sep-06 19:11

Thanks Steve.

BlitzPackage29-Sep-06 17:23

29-Sep-06 17:23

Greetings Colleagues,

I have a quick question. How do you return an object by value?

The following code:

<br />
MyObject MyClass::operator+(const MyClass & mcRhs)<br />
{<br />
     //...do some processing<br />
     MyClass mcNewObject();<br />
     return mcNewObject;<br />
}<br />

results in the following error: c++ error C2664: No constructor could take the source type, or constructor overload resolution was ambiguous

It says it can't convert it. Ironically, when I leave the parentheses off of the instantiation of mcNewObject, it works!? Can someone please explain.

Thanks a million.

Sincerely,
BP

Mike_V29-Sep-06 17:36

Mike_V

29-Sep-06 17:36

Can you post the definitions of the two classes?

Or... what line was the error on?

Mike

BlitzPackage29-Sep-06 17:43

29-Sep-06 17:43

Thanks for replying,

This is the actual function

<br />
CNumber CNumber::operator+(const CNumber & cnThisNumber)<br />
{<br />
	if((m_bIsBlank) || (cnThisNumber.ReturnIsBlank()))<br />
	{<br />
		CNumber cnBlankNumber();<br />
		return cnBlankNumber;<br />
	}<br />
	else<br />
	{<br />
		float fReturnAmount;<br />
		cnThisNumber.GetAmount(fReturnAmount);<br />
		fReturnAmount = m_fAmount - fReturnAmount;<br />
		CNumber cnNewAmount(fReturnAmount);<br />
		return cnNewAmount;<br />
	}<br />
}<br />

This is the copy constructor I have defined...

<br />
CNumber::CNumber(const CNumber & cnThisNumber) : m_iMaxDecimals(10)<br />
{<br />
	// Set all changeable data to cnThisNumber<br />
	cnThisNumber.GetAmount(m_fAmount);<br />
	cnThisNumber.GetOriginalAmount(m_fOriginalAmount);<br />
	m_bFormatChange = cnThisNumber.ReturnFormatChange();<br />
	m_bIsZero = cnThisNumber.ReturnIsZero();<br />
	m_strStringAmount = cnThisNumber.ReturnStringAmount();<br />
	m_bIsBlank = cnThisNumber.ReturnIsBlank();<br />
}<br />

If I change "CNumber cnBlankNumber()" in the operator overload to "CNumber cnBlankNumber" it works. I just don't get it.

Mike_V29-Sep-06 17:59

Mike_V

29-Sep-06 17:59

I don't get the () thing, but try doing return CNumber; or return CNumber();.

Also - do you have a default constructor defined?

Mike

BlitzPackage29-Sep-06 18:04

29-Sep-06 18:04

I will try that, and yes I do have a default constructor defined.

Thanks,
BP