How do i round of a float? say i have an integer x which is 501 but x must be a power of 2. so: float fPower = log10(x) / log10(2); x = pow(2, fPower); fPower should be rounded off to an integer, how do i do this? Now when fPower is e.g. 8.9 it is used as 8 instead of 9 ---------- Drago

Hey, try tyepcasting the float...if x is an int, fPower will be read into the equation as an intiger...

i know this is simple, and you probalby know better, but here's a little better solution. I dont know of any built in functions, but you could make a simple function to round up for down...like:

int roundNum(float fltNum)
{
int intNum;
float difference;

int intNum = fltNum;

difference = fltNum - (float)intNum;

if(difference >= 0.5)
intNum++;

return intNum;
}



OK, try that...real simple and bare-bones basic rounding function...i did it in a simple test program and it worked without fail (though you might want to change int to a larger variable type if needed)

Email me if it works, or if it doesn't, i'd like to know if i actually helped and i dont come here often...
roosterjm@home.com

Edited by - roosterjm on 4/29/00 10:53:42 PM

Why dun you just think simple ?

int ConvFloat(float a)
{
return( (int)(a+0.5)) ;

}

U see ... if the decimal part is anything less than 0.5, adding 0.5 to a will not add to the interger part of the float. Therefore will be trancated and thus rounded down. If the decimal part is 0.5 or more, the interger part will increase after adding 0.5. After trancation, the float will be rounded up

typecasting int doesn''t round the float, it just truncates the decimal part . To do this, you need your own function. The last guy''s function is pretty good, although i would delare it inline since it would be used so often .

Dont you think thats fast enough? it should be...

========================
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www.fiend.cjb.net

Yup, the "int (a + 0.5)" is a classic trick that works great.

Tim

Just a quick note: the (int)(a+0.5) trick only works for positive numbers. If the number is negative you need (int)(a-0.5).

Edited by - Ranok on 5/4/00 6:54:27 PM

if you wanted pos and neg just have
int (a + 0.5*((a<0)? -1:1))
erm...i think that''s right

hmm .... how abt this

int convStuff( float a)
{
float z = abs(a) ;
int b = int(a/z) ; //so its either -1 or 1, depending on a
return (int(z+0.5)*b) ;
}

haha just do

return (int)(flt + flt<0 ? -0.5 : 0.5);

no fancy stuff needed!


Edited by - adammil on May 9, 2000 2:13:07 PM