Thomas Krog Christensen wrote:

> > Is this compiler dependent?  Don't C++ programs truncate when converting
> > from double to int?
>
> Yes they do. But if you want to truncate from double to double I think it's
> faster to use floor.(if not there won't be any reason to use floor)

Sure there would.  What happens when you try to round a number such as 1E50 by
casting to an int?  A typical signed 32 bit integer can only hold numbers on
the order of 10^10.  So an advantage of floor is that it works for all possible
values of double rather than just the small ones.

Note: I'm not saying floor isn't also faster than casting to int.  It may be.
I'm saying it has a purpose regardless of its speed.

Finally, let me point out that this also presents a problem for the RoundOff
function proposed earlier:

> double void RoundOff(double x,int noDecimal){
>     int f=x*pow(10,noDecimal);
>     return floor(x*f+0.5)/f;
> }

I believe a better solution would be to use the modf function (in <math.h>, or
since this is comp.std.c++, <cmath>).  Quoting from K&R2, "modf(double x,
double *ip) splits x into integral and fractional parts, each with the same
sign as x.  It stores the integral part in *ip, and returns the fractional
part."

-----
Gillmer J. Derge

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Paul Meyerholtz wrote:
> A question about the function floor() from a newbie.  If I'm trying to
> round off doubles the following seems to work:
>
> double x;
> ....   x assigned some number ....
> int rounded = int(x +.5);
>
> Is this compiler dependent?  Don't C++ programs truncate when converting
> from double to int?


You should note that using floor() is different from truncating when the
value is negative, e.g. floor(-2.4) is -3 while int(-2.4) is -2.

- Kaspar

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Thomas Krog Christensen wrote:
>
> "Tomas Alvier" <tal@nfis.no> wrote in message
> news:C5Ns4.5456$kz6.102045@news1.online.no...
> > Is there a function for rounding of a double precision variable to a
> chosen
> > number of decimals?
>
> double void RoundOff(double x,int noDecimal){
>     int f=x*pow(10,noDecimal);
>     return floor(x*f+0.5)/f;
> }

A question about the function floor() from a newbie.  If I'm trying to
round off doubles the following seems to work:

double x;
....   x assigned some number ....
int rounded = int(x +.5);

Is this compiler dependent?  Don't C++ programs truncate when converting
from double to int?

Paul

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> Is this compiler dependent?  Don't C++ programs truncate when converting
> from double to int?

Yes they do. But if you want to truncate from double to double I think it's
faster to use floor.(if not there won't be any reason to use floor)


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Thomas Krog Christensen wrote:
>
> "Tomas Alvier" <tal@nfis.no> wrote in message
> news:C5Ns4.5456$kz6.102045@news1.online.no...
> > Is there a function for rounding of a double precision variable to a
> chosen
> > number of decimals?
>
> double void RoundOff(double x,int noDecimal){
>     int f=x*pow(10,noDecimal);
>     return floor(x*f+0.5)/f;
> }
>
Of course computationally that will do it, but the real question is
what the original poster really wanted.  If he wanted the value only
"DISPLAYED" to a certain number of decimal places, this won't accomplish
it.

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Is there a function for rounding of a double precision variable to a chosen
number of decimals?

Thanks in advance,
Tomas Alvier


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"Tomas Alvier" <tal@nfis.no> wrote in message
news:C5Ns4.5456$kz6.102045@news1.online.no...
> Is there a function for rounding of a double precision variable to a
chosen
> number of decimals?

double void RoundOff(double x,int noDecimal){
    int f=x*pow(10,noDecimal);
    return floor(x*f+0.5)/f;
}


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Ron Natalie wrote:
>Of course computationally that will do it, but the real question is
>what the original poster really wanted.  If he wanted the value only
>"DISPLAYED" to a certain number of decimal places, this won't accomplish
>it.

then you can use:

char buffer[100];
double v=5.457;
sprintf(buffer,"%.2f",v);

(but I don't think this is the answer to the original qustion)


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