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Round Function

Posted on 2001-08-11
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Last Modified: 2008-03-03
I am currently programming in a language called Visual Dataflex.    It can get the job done, but it lacks many of the creature comforts of C++ and VB.

There is on round decent round function and there is no Array type.  there is an Array class, but the syntax for using it and most other VDF stuff is very long and clunky.

Hence my problem.  I want to create my own Round function.  Back in C++, I made one with a CHAR and it was pretty easy.  This function will only need to round things after the decimal place.  Everything I've tried so far is very long and I was wondering if anyone can suggest an algorithm for rounding after the decimal point, assuming that I can't use an array.

Essentially, I want the function to take two arguments, the number and the number to the right of the decimal place (Just like in other languages).

I know this is painfully easy, but I'm having a block.

Thanks,

Bill

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Question by:wgrmba
13 Comments
 
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Expert Comment

by:Triskelion
ID: 6376242
>>This function will only need to round things after the decimal place.  

Can you simply add the logic of:
   If the number after the decimal is greater than or equal to 50 then add one to the number before the decimal and return it?
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Expert Comment

by:MatthewL
ID: 6376736
Hi wgrmba,

You can add .5 to the number and then cast to an integer.

To handle rounding to the 1 decimal place, multiply by 10 add .5, cast, divide by 10

To handle rounding to the nth decimal place, muliply by 10^n, add .5, cast, divide by 10^n

Matt
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Expert Comment

by:TrueDrake
ID: 6377963
Hi,

z = 1.56
x = Round(z, 1)
MsgBox x

Enjoy
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Expert Comment

by:t0t0
ID: 6378281

integer = ABS(float)

IF (float - integer >= 0.5)
  integer = integer + 1
END-IF

or in C


.
.
i = abs(f);

if (f - i >= 0.5)
  i++;

.
.


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Expert Comment

by:griessh
ID: 6382183
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by:griessh
ID: 6382188
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by:griessh
ID: 6382199
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by:griessh
ID: 6382201
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by:griessh
ID: 6382205
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Expert Comment

by:griessh
ID: 6382724
I can't post anymore ... *&@*&$*&)(&(*^%!!!!
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Accepted Solution

by:
griessh earned 100 total points
ID: 6382726
AHHHH! It works again. Sorry, but I never got anything through to EE. So here you go:
Since I don't know the syntax of your language, here a very general function for rounding to a specific decimal. It is really just to give you the idea...

float round (float f, int dec)
{
  int k =1;
  for (i=0; i<dec; i++) { // Multiply by dec 10s
     f=f*10;
     k=k*10;
  }
  // Rounding is +.5 then take int
  f = f+0.5;
  int res = int (f);

  // Make it a float again
  float ret = ((float) res) / k;

  return (ret);
}

======
Werner
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Expert Comment

by:ozo
ID: 6385806
That doesn't work for negative numbers.
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Expert Comment

by:griessh
ID: 6385860
ozo

Thanks for the comment. As I said: "It is really just to give you the idea...".

There is NO checking of any kind in this function! But I guess that was not asked for anyhow ...

======
Werner
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