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C++ round to n places

Posted on 2006-11-20
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1,413 Views
Last Modified: 2008-01-09
Hi experts,
I have two related questions:

(A)
I would like to know if there is a fuction to round double values to n decimal places e.g.

double value = 0.000000000234567897

value = round(value, 12) // now value is 0.000000000235 i.e to 12 decimal places

(B)
How do I watch a double value to n decimal places in visual studio c++. When I hover over a double variable I can only see up to 6 or 7 decimal places?

Thank you,
j

0
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Question by:XPUSR
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9 Comments
 
LVL 22

Expert Comment

by:grg99
ID: 17982189

you could multiply the number by 10^places, truncate it to an integer, then divide by 10^places.

0
 
LVL 7

Expert Comment

by:UrosVidojevic
ID: 17982370
or you could include the following:

#include <iostream>
#include <iomanip>
using namespace std;

and then use manipulator setprecision(int t);

which will round your number at first t-1 decimal positions.

for example:

double x = 1.23456789;
cout << setprecision(4) << x;

will print 1.234.

Howerever, when numbers are small like your double value, then compiler will
use exponential notation for example 2.346 e-010 (in this case), where
"first number" is rounded at t-1 decimal places.

so:

double value = 0.00000000023456789;
cout << setprecision(4) << value;

will print 2.346 e-010.
0
 
LVL 7

Expert Comment

by:UrosVidojevic
ID: 17982424
Sorry about a little mistake, in those t positions are counted digits both after and before point.
You can see complete explanation here.

http://www.cplusplus.com/ref/iostream/iomanip/setprecision.html
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Author Comment

by:XPUSR
ID: 17982769
Thanks very much but I cannot use cout as I am in a COM environment. Is there another way?


Also,

double TeamAScore1 = 7.00;
double TeamAScore2 = 7.02;
double performace = (TeamAScore2 /TeamAScore1) - 1;

If I debug the code above I get a value of "0.0028571428571429" in the watch window for the performance variable, however if I do the following:

double endperformace = performace - 0.0028571428571429;

the endperformace variable is not 0 but "-8.6736173798840e-018". Can anyone explain this?

0
 

Author Comment

by:XPUSR
ID: 17983033
Is there a mehod of outputing data to n decimal places to the visual c++ output window?
0
 
LVL 8

Expert Comment

by:YoungBonzi
ID: 17984416
Hello, give this a shot:

                         double value = 0.000000000234567897;
                         double rounded;
                         int n = 12;
                         rounded = System::Math::Round(value, n);
                         MessageBox::Show("value: "+value+"\nrounded: "+rounded);

For Round() use the 4th function of it (4 of 8). The default will round to the nearest integer. The 4th rounds to desired precision.
0
 
LVL 8

Expert Comment

by:YoungBonzi
ID: 17984420
0
 

Author Comment

by:XPUSR
ID: 17985340
>> YoungBonzi
I think the System::Math::Round is a .NET method. I can't seem to get it to work for non .Net apps?
0
 
LVL 4

Accepted Solution

by:
bdunz19 earned 130 total points
ID: 18001371
For the rounding:

float RoundUp(float Value, int Places)
{
      float tmp = 0.0;
      int hold = 0;
      hold = (int)(Value * (float)pow(10, Places) + 0.5);
      tmp = (float)hold / (float)pow(10, Places);

      return tmp;
}

As for long numbers, the best I can come up with is to store your entire number in an advanced double array. Meaning each increment in the array stores the value of data to a further percision (or maybe some simple short array with one single digit value for each percision).
0

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