rounding a float/double to nearest 1/10th

[Shea Martin]

Any one have a better/simpler method for rounding a float to the nearest
1/10th? This is currently what I am using, and there must be a better
way, or perhaps a canned method that I am not aware of.

double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;

NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.

Any suggestions would be great thank.

~S

[Shea Martin]

Shea Martin wrote:

Any one have a better/simpler method for rounding a float to the nearest
1/10th? This is currently what I am using, and there must be a better
way, or perhaps a canned method that I am not aware of.

double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;

NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.

Any suggestions would be great thank.

~S

float z2 = atof(arg[1]);
z2 = (float)floor(z2*10+0.5)/10;

I thinks this is a lot better.

~S

[Oliver S.]

> float z2 = atof(arg[1]);

z2 = (float)floor(z2*10+0.5)/10;

better

z2 = (float)(floor( z2 * 10 + 0.5 ) / 10)

So the value gets chopped to a float (if at all) after the division.

[Gianni Mariani]

Shea Martin wrote:

Any one have a better/simpler method for rounding a float to the nearest
1/10th? This is currently what I am using, and there must be a better
way, or perhaps a canned method that I am not aware of.

double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;

NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.

z = 0.1 * round( z * 10.0 );

[Shea Martin]

Gianni Mariani wrote:

Shea Martin wrote:

Any one have a better/simpler method for rounding a float to the
nearest 1/10th? This is currently what I am using, and there must be
a better way, or perhaps a canned method that I am not aware of.

double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;

NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.

z = 0.1 * round( z * 10.0 );

Which header do I need to get round? It does not seem to be in math.h
on Solaris 9.

Thanks,

~S

[Chris Theis]

"Gianni Mariani" <gi*******@mariani.ws> wrote in message
news:bn********@dispatch.concentric.net...
[SNIP]

z = 0.1 * round( z * 10.0 );

Somehow I missed that round() was a standard function. I'd be happy if you
could point out where I can find it, so that I can get rid of my own
solution.

I usually use the following approach:

double Round( double Value, int Digits )
{
if( Value > 0.0 )
return ( (long)( Value * Faktor + 0.5 ) ) / pow( 10.0, Digits);

return ( (long)( Value * Faktor - 0.5 ) ) / pow( 10.0, Digits);
}

In case of common values for Digits one could use a table instead of
calculating the factor with pow() as it is a rather slow function.

Regards
Chris

[Chris Theis]

"Chris Theis" <Ch*************@nospam.cern.ch> wrote in message
news:bn**********@sunnews.cern.ch...

"Gianni Mariani" <gi*******@mariani.ws> wrote in message
news:bn********@dispatch.concentric.net... [SNIP] double Round( double Value, int Digits )
{
if( Value > 0.0 )
return ( (long)( Value * pow( 10.0, Digits) + 0.5 ) ) / pow( 10.0, Digits);

Sorry, this should of course be
if( Value > 0.0 )
return ( (long)( Value * pow( 10.0, Digits) + 0.5 ) ) / pow( 10.0,
Digits);

return ( (long)( Value * pow( 10.0, Digits) - 0.5 ) ) / pow( 10.0,
Digits);
}

It's obviously too early for me :-)

Chris

[Ron Natalie]

"Shea Martin" <sm*****@arcis.com> wrote in message news:fNCnb.9864$f7.536027@localhost...

z = 0.1 * round( z * 10.0 );

Which header do I need to get round? It does not seem to be in math.h
on Solaris 9.

Ain't no such function. Try nearbyint or rint.

[Ron Natalie]

"Chris Theis" <Ch*************@nospam.cern.ch> wrote in message news:bn**********@sunnews.cern.ch...

"Gianni Mariani" <gi*******@mariani.ws> wrote in message
news:bn********@dispatch.concentric.net...
[SNIP]

z = 0.1 * round( z * 10.0 );

Somehow I missed that round() was a standard function. I'd be happy if you
could point out where I can find it, so that I can get rid of my own
solution.

Should be in math.h. It's a C function, but part of C99.

[Chris Theis]

"Ron Natalie" <ro*@sensor.com> wrote in message
news:3f***********************@news.newshosting.co m...

"Chris Theis" <Ch*************@nospam.cern.ch> wrote in message

news:bn**********@sunnews.cern.ch...

"Gianni Mariani" <gi*******@mariani.ws> wrote in message
news:bn********@dispatch.concentric.net...
[SNIP]

z = 0.1 * round( z * 10.0 );

Somehow I missed that round() was a standard function. I'd be happy if you could point out where I can find it, so that I can get rid of my own
solution.

Should be in math.h. It's a C function, but part of C99.

Aah, thanks Ron. Didn't know that.

Cheers
Chris

[Shea Martin]

Ron Natalie wrote:

"Chris Theis" <Ch*************@nospam.cern.ch> wrote in message news:bn**********@sunnews.cern.ch...

"Gianni Mariani" <gi*******@mariani.ws> wrote in message
news:bn********@dispatch.concentric.net...
[SNIP]

z = 0.1 * round( z * 10.0 );

Somehow I missed that round() was a standard function. I'd be happy if you
could point out where I can find it, so that I can get rid of my own
solution.

Should be in math.h. It's a C function, but part of C99.

One more reason to upgrade our aging compilers (Sun WorkShop 5.0). :-)
~S

[Anonymous]

Shea Martin wrote:

Any one have a better/simpler method for rounding a float to the nearest
1/10th?

Don't know what you want to do with the result, but how about

char s[30];
sprintf(s,"%.1lf",z); // s is z rounded to nearest 0.1
z=atof(s);

sh************@computer.org (remove caps for e-mail)

[Steven C.]

"Shea Martin" <sm*****@arcis.com> wrote in message
news:fNCnb.9864$f7.536027@localhost...
Gianni Mariani wrote:

Shea Martin wrote:

Any one have a better/simpler method for rounding a float to the
nearest 1/10th? This is currently what I am using, and there must be
a better way, or perhaps a canned method that I am not aware of.

double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;

NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.

z = 0.1 * round( z * 10.0 );

Which header do I need to get round? It does not seem to be in math.h
on Solaris 9.

Thanks,

~S

----------------------------------------------
Personally I prefer a car.

[P.J. Plauger]

"Steven C." <no****@xxx.com> wrote in message news:oW*****************@twister.socal.rr.com...

"Shea Martin" <sm*****@arcis.com> wrote in message
news:fNCnb.9864$f7.536027@localhost...
Gianni Mariani wrote:

Shea Martin wrote:

Any one have a better/simpler method for rounding a float to the
nearest 1/10th? This is currently what I am using, and there must be
a better way, or perhaps a canned method that I am not aware of.

double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;

NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.

z = 0.1 * round( z * 10.0 );

Which header do I need to get round? It does not seem to be in math.h
on Solaris 9.

You need a version of math.h that better conforms to C99. We offer
such a library, but it's an extra-cost item.

P.J. Plauger
Dinkumware, Ltd.
http://www.dinkumware.com