Rounding a floating point number

[jacob navia]

Hi

"How can I round a number to x decimal places" ?

This question keeps appearing. I would propose the following
solution

#include <float.h>
#include <math.h>

double roundto(double x, int digits)
{
int sgn=1;

if (x == 0)
return 0;
if (digits DBL_DIG)
digits = DBL_DIG;
else if (digits < 1)
digits = 1;

if(x < 0.0) {
sgn = -sgn;
x = -x;
}
double p = floorl(log10l(x ));
p--;
p = digits-p;
double pow10 = pow(10.0, p);
return sgn*floor(x*pow 10+0.5)/pow10;
}

long double roundtol(long double x, int digits)
{
int sgn=1;

if (x == 0)
return 0;
if (digits LDBL_DIG)
digits = LDBL_DIG;
else if (digits < 1)
digits = 1;

if(x < 0.0) {
sgn = -sgn;
x = -x;
}
long double p = floorl(log10l(x ));
p--;
p = digits-p;
long double pow10 = powl(10.0, p);
return sgn*floorl(x*po w10+0.5)/pow10;
}
#include <stdio.h>
int main(void)
{
double d = 1.7888889988996 678;
long double ld = 1.7888889988996 678998877L;

for (int i = 0; i<=DBL_DIG;i++ ) {
printf("i=%d: %.15g\n",i,roun dto(d,i));
}
printf("\n 1.7888889988996 678\n");
for (int i = 0; i<=LDBL_DIG;i++ ) {
printf("i=%d: %.18Lg\n",i,rou ndtol(ld,i));
}
printf("\n 1.7888889988996 678998877L\n");
return 0;
}

--------------------------------------------------------------------
I would propose it to add it to the FAQ.
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[Richard Heathfield]

jacob navia said:

Hi

"How can I round a number to x decimal places" ?

This question keeps appearing.

<snip>

I would propose it to add it to the FAQ.

Good idea. You could call it Question 14.6.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Bartc]

"Richard Heathfield" <rj*@see.sig.in validwrote in message
news:xc******** *************** *******@bt.com. ..

jacob navia said:

>Hi

"How can I round a number to x decimal places" ?

This question keeps appearing.

<snip>

>I would propose it to add it to the FAQ.

Good idea. You could call it Question 14.6.

14.6 is already used by How to Round a Number [to an integer]

--
Bart

[jacob navia]

Richard Heathfield wrote:

jacob navia said:

>Hi

"How can I round a number to x decimal places" ?

This question keeps appearing.

<snip>

>I would propose it to add it to the FAQ.

Good idea. You could call it Question 14.6.

Your kill file is not working apparently
:-)

That question addresses the rounding to nearest integer.
This answers the question of rounding to a number of decimal places

Not the same thing. But yes, we could add this to 14.6

--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[MisterE]

14.6 is already used by How to Round a Number [to an integer]

Then what need is there for this question? Its already answered in the faq.
Multiply, round then divide.

[Richard Heathfield]

Bartc said:

>
"Richard Heathfield" <rj*@see.sig.in validwrote in message
news:xc******** *************** *******@bt.com. ..

>jacob navia said:

>>Hi

"How can I round a number to x decimal places" ?

This question keeps appearing.

<snip>

>>I would propose it to add it to the FAQ.

Good idea. You could call it Question 14.6.

14.6 is already used by How to Round a Number [to an integer]

From the FAQ, Q14.6: "You can round to a certain precision by scaling:

(int)(x / precision + 0.5) * precision"

(int)(3.141 / 0.01 + 0.5) * 0.01 gives us:

(int)(314.1 + 0.5) * 0.01 which is

(int)314.6 * 0.01 which is

314 * 0.01 which is

3.14

QED

(Of course, the usual caveats about floating point apply here.)

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Martien Verbruggen]

On Mon, 25 Feb 2008 12:19:35 +0100,
jacob navia <ja***@nospam.c omwrote:

Richard Heathfield wrote:

>jacob navia said:

>>Hi

"How can I round a number to x decimal places" ?

This question keeps appearing.

<snip>

>>I would propose it to add it to the FAQ.

Good idea. You could call it Question 14.6.

That question addresses the rounding to nearest integer.
This answers the question of rounding to a number of decimal places

It also answers the question of how to round to a certain number of
digits:

\begin{quote}
You can round to a certain precision by scaling:
\end{quote}

Martien
--
|
Martien Verbruggen | prepBut nI vrbLike adjHungarian! qWhat's
| artThe adjBig nProblem? -- Alec Flett
|

[Bartc]

"Richard Heathfield" <rj*@see.sig.in validwrote in message
news:oZ******** *************@b t.com...

Bartc said:

>>
"Richard Heathfield" <rj*@see.sig.in validwrote in message
news:xc******* *************** ********@bt.com ...

>>jacob navia said:

Hi

"How can I round a number to x decimal places" ?

This question keeps appearing.

<snip>

I would propose it to add it to the FAQ.

Good idea. You could call it Question 14.6.

14.6 is already used by How to Round a Number [to an integer]

From the FAQ, Q14.6: "You can round to a certain precision by scaling:

(int)(x / precision + 0.5) * precision"

(int)(3.141 / 0.01 + 0.5) * 0.01 gives us:

(int)(314.1 + 0.5) * 0.01 which is

(int)314.6 * 0.01 which is

314 * 0.01 which is

3.14

Yeah, you're right. Perhaps 14.6 should be written more clearly, with regard
to what exactly it means by Precision. And an example added, like yours.

>(int)(3.141 / 0.01 + 0.5) * 0.01

Somehow I would be happier if that was written as:

>(int)(3.141 *100 + 0.5) /100

Then any errors due to 0.01 not being exactly 1/100 are someone else's
fault.

--
Bart

[Mark McIntyre]

MisterE wrote:

>14.6 is already used by How to Round a Number [to an integer]

Then what need is there for this question?

<Fx: swoosh sound of clue going over head>

--
Mark McIntyre

CLC FAQ <http://c-faq.com/>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt >

[user923005]

On Feb 25, 2:18*am, jacob navia <ja...@nospam.c omwrote:

Hi

"How can I round a number to x decimal places" ?

This question keeps appearing. I would propose the following
solution

#include <float.h>
#include <math.h>

double roundto(double x, int digits)
{
* * * * int sgn=1;

* * * * if (x == 0)
* * * * * * * * return 0;
* * * * if (digits DBL_DIG)
* * * * * * * * digits = DBL_DIG;
* * * * else if (digits < 1)
* * * * * * * * digits = 1;

* * * * if(x < 0.0) {
* * * * * * * * sgn = -sgn;
* * * * * * * * x = -x;
* * * * }
* * * * double p = floorl(log10l(x ));
* * * * p--;
* * * * p = digits-p;
* * * * double pow10 = pow(10.0, p);
* * * * return sgn*floor(x*pow 10+0.5)/pow10;

}

long double roundtol(long double x, int digits)
{
* * * * int sgn=1;

* * * * if (x == 0)
* * * * * * * * return 0;
* * * * if (digits LDBL_DIG)
* * * * * * * * digits = LDBL_DIG;
* * * * else if (digits < 1)
* * * * * * * * digits = 1;

* * * * if(x < 0.0) {
* * * * * * * * sgn = -sgn;
* * * * * * * * x = -x;
* * * * }
* * * * long double p = floorl(log10l(x ));
* * * * p--;
* * * * p = digits-p;
* * * * long double pow10 = powl(10.0, p);
* * * * return sgn*floorl(x*po w10+0.5)/pow10;}

#include <stdio.h>
int main(void)
{
* * * * double d = 1.7888889988996 678;
* * * * long double ld = 1.7888889988996 678998877L;

* * * * for (int i = 0; i<=DBL_DIG;i++ ) {
* * * * * * * * printf("i=%d: %.15g\n",i,roun dto(d,i));
* * * * }
* * * * printf("\n * * *1.788888998899 6678\n");
* * * * for (int i = 0; i<=LDBL_DIG;i++ ) {
* * * * * * * * printf("i=%d: %.18Lg\n",i,rou ndtol(ld,i));
* * * * }
* * * * printf("\n * * *1.788888998899 6678998877L\n") ;
* * * * return 0;

}

--------------------------------------------------------------------
I would propose it to add it to the FAQ.

This is the snippet's solution (which is very similar to yours):
/* round number n to d decimal points */
double fround(double n, unsigned d)
{
return floor(n * pow(10., d) + .5) / pow(10., d);
}

Of course, no solution is really going to be satisfying, as long as
the output format is floating point.
Another question is "What kind of rounding?" Do we want banker's
rounding or round to nearest or what?
Should we create a round function for each distinct floating point
type?