C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  rounding a float/double to nearest 1/10th          C++Talk.NET Forum Index -> C++ language (comp.lang.c++) View previous topic :: View next topic   Author Message Shea Martin Guest Posted: Tue Oct 28, 2003 10:14 pm    Post subject: rounding a float/double to nearest 1/10th 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 Back to top Shea Martin Guest Posted: Tue Oct 28, 2003 10:32 pm    Post subject: Re: rounding a float/double to nearest 1/10th Shea Martin wrote: Quote: 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 Back to top Gianni Mariani Guest Posted: Tue Oct 28, 2003 10:37 pm    Post subject: Re: rounding a float/double to nearest 1/10th Shea Martin wrote: Quote: 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 ); Back to top Shea Martin Guest Posted: Tue Oct 28, 2003 11:13 pm    Post subject: Re: rounding a float/double to nearest 1/10th Gianni Mariani wrote: Quote: 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 Back to top Chris Theis Guest Posted: Wed Oct 29, 2003 9:58 am    Post subject: Re: rounding a float/double to nearest 1/10th "Gianni Mariani" <gi2nospam (AT) mariani (DOT) ws> wrote [SNIP] Quote: 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 Back to top Chris Theis Guest Posted: Wed Oct 29, 2003 10:04 am    Post subject: Correction Re: rounding a float/double to nearest 1/10th "Chris Theis" <Christian.Theis (AT) nospam (DOT) cern.ch> wrote Quote: "Gianni Mariani" <gi2nospam (AT) mariani (DOT) ws> wrote in message news:bnmr3c$j3u (AT) dispatch (DOT) concentric.net... [SNIP] double Round( double Value, int Digits ) { if( Value > 0.0 ) return ( (long)( Value * pow( 10.0, Digits) + 0.5 ) ) / pow( 10.0, Digits); Sorry, this should of course be if( Value > 0.0 ) return ( (long)( Value * pow( 10.0, Digits) + 0.5 ) ) / pow( 10.0, Digits); return ( (long)( Value * pow( 10.0, Digits) - 0.5 ) ) / pow( 10.0, Digits); Quote: } It's obviously too early for me :-) Chris Back to top Ron Natalie Guest Posted: Wed Oct 29, 2003 4:24 pm    Post subject: Re: rounding a float/double to nearest 1/10th "Shea Martin" <smartin (AT) arcis (DOT) com> wrote Quote: z = 0.1 * round( z * 10.0 ); Which header do I need to get round? It does not seem to be in math.h on Solaris 9. Ain't no such function. Try nearbyint or rint. Back to top Ron Natalie Guest Posted: Wed Oct 29, 2003 4:26 pm    Post subject: Re: rounding a float/double to nearest 1/10th "Chris Theis" <Christian.Theis (AT) nospam (DOT) cern.ch> wrote Quote: "Gianni Mariani" <gi2nospam (AT) mariani (DOT) ws> wrote in message news:bnmr3c$j3u (AT) dispatch (DOT) 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. Back to top Chris Theis Guest Posted: Wed Oct 29, 2003 4:41 pm    Post subject: Re: rounding a float/double to nearest 1/10th "Ron Natalie" <ron (AT) sensor (DOT) com> wrote Quote: "Chris Theis" <Christian.Theis (AT) nospam (DOT) cern.ch> wrote "Gianni Mariani" <gi2nospam (AT) mariani (DOT) ws> wrote in message news:bnmr3c$j3u (AT) dispatch (DOT) concentric.net... [SNIP] z = 0.1 * round( z * 10.0 ); Somehow I missed that round() was a standard function. I'd be happy if you could point out where I can find it, so that I can get rid of my own solution. Should be in math.h. It's a C function, but part of C99. Aah, thanks Ron. Didn't know that. Cheers Chris Back to top Shea Martin Guest Posted: Wed Oct 29, 2003 8:20 pm    Post subject: Re: rounding a float/double to nearest 1/10th Ron Natalie wrote: Quote: "Chris Theis" <Christian.Theis (AT) nospam (DOT) cern.ch> wrote "Gianni Mariani" <gi2nospam (AT) mariani (DOT) ws> wrote in message news:bnmr3c$j3u (AT) dispatch (DOT) 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 Back to top Anonymous Guest Posted: Thu Oct 30, 2003 2:42 am    Post subject: Re: rounding a float/double to nearest 1/10th Shea Martin wrote: Quote: 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); [email]sherNOwoodSPAM (AT) computer (DOT) org[/email] (remove caps for e-mail) Back to top Steven C. Guest Posted: Thu Oct 30, 2003 6:09 am    Post subject: Re: rounding a float/double to nearest 1/10th "Shea Martin" <smartin (AT) arcis (DOT) com> wrote Gianni Mariani wrote: Quote: 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. Back to top P.J. Plauger Guest Posted: Thu Oct 30, 2003 4:51 pm    Post subject: Re: rounding a float/double to nearest 1/10th "Steven C." <nospam (AT) xxx (DOT) com> wrote Quote: "Shea Martin" <smartin (AT) arcis (DOT) com> wrote in message news:fNCnb.9864$f7.536027 (AT) localhost (DOT) .. 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 Back to top Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First         C++Talk.NET Forum Index -> C++ language (comp.lang.c++) All times are GMT Page 1 of 1   Jump to: Select a forum C/C++----------------C LanguageC Language (Moderated)C++ Language (Moderated)C++ language (comp.lang.c++)C++ language, library and standards Non-English C/C++ Newsgroups----------------C++ (German)C++ (French)  You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum Powered by phpBB © 2001, 2006 phpBB Group SEO toolkit © 2004-2006 webmedic.