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View Full Version : ceil - floor is ok.but where is round ?

 claudiof03-11-2009, 02:34 PMis there any function can : round(3.6) //result 4 round(3.4) //result 3 ?
DOor
03-11-2009, 02:59 PM
I know I have had this issue before but for the life of me cant remember if I found a command to achieve this.

off the top of my head I know you could get the decimal by using % and then if the decimal is greater than .5 you could Ceil the number, if its less than .5 you could Floor it.

the below code will give you the results you need. note the % 1 that i did-- that is dividing by 1 and getting the remainder (so you get only the decimal of the float) then I just check if it is above or below 0.5.

float \$myNumber = 3.2;
float \$n = \$myNumber % 1;

if (\$n > 0.5) {
\$myNumber = `ceil \$myNumber`;
} else {
\$myNumber = `floor \$myNumber`;
}

print \$myNumber;

YourDaftPunk
03-11-2009, 10:33 PM
Try adding 0.5 and calling floor. Drop it in a function callled round() and you are set:

global proc float round(float \$x) {
return floor(\$x + 0.5);
}

// some float number
float \$random = rand(0,10);

// round to int
\$rounded = round(\$random);

DOor
03-11-2009, 10:42 PM
ah nice, the +.5 method is definitely more efficient. I knew there was a better way, couldnt think of anything off the top of my head. :shrug:

dbiggs
03-12-2009, 02:17 PM
Here's a more complex version for rounding to n decimal places.

//
// Roundoff a float to \$n places.
//
global proc float roundoff( float \$f, int \$n )
{
// we divide if n < 0 to avoid numeric
// precision problems
if( \$n > 0 )
{
float \$roundScale = pow(10,\$n);
if( \$f > 0 )
return( ((float)(int)(\$f * \$roundScale + 0.5)) /\$roundScale );
else
return( ((float)(int)(\$f * \$roundScale - 0.5)) /\$roundScale );
}
else
{
float \$roundScale = pow(10,-\$n);
if( \$f > 0 )
return( ((float)(int)(\$f/\$roundScale + 0.5)) *\$roundScale );
else
return( ((float)(int)(\$f/\$roundScale - 0.5)) *\$roundScale );

}
}

berniebernie
03-12-2009, 05:57 PM
i think there's a trick with INTs: http://www.foreachloop.com/blog/?p=7

or

it's a little silly that it's not included as a standard function

edit: did not read post above mine =)

acidream
03-12-2009, 11:49 PM
heres how i round numbers:

(floor(number_to_round * 10) / 10)

if the number is 44.53256222 it will return 44.5

change the 10 to 100 and it will return 44.53

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03-12-2009, 11:49 PM
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