Date: Mon, 13 Aug 2012 15:39:24 -0500 From: Stephen Montgomery-Smith <stephen@missouri.edu> To: Steve Kargl <sgk@troutmask.apl.washington.edu> Cc: freebsd-numerics@freebsd.org, Bruce Evans <brde@optusnet.com.au> Subject: Re: Complex arg-trig functions Message-ID: <502965FC.1040203@missouri.edu> In-Reply-To: <20120813201626.GA54144@troutmask.apl.washington.edu> References: <20120805030609.R3101@besplex.bde.org> <501D9C36.2040207@missouri.edu> <20120805175106.X3574@besplex.bde.org> <501EC015.3000808@missouri.edu> <20120805191954.GA50379@troutmask.apl.washington.edu> <20120807205725.GA10572@server.rulingia.com> <20120809025220.N4114@besplex.bde.org> <5027F07E.9060409@missouri.edu> <20120814003614.H3692@besplex.bde.org> <50295887.2010608@missouri.edu> <20120813201626.GA54144@troutmask.apl.washington.edu>
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On 08/13/2012 03:16 PM, Steve Kargl wrote: > On Mon, Aug 13, 2012 at 02:41:59PM -0500, Stephen Montgomery-Smith wrote: >> >> Also, you made the comment that in the float version, all the 0.5 should >> become 0.5F. Two questions: >> 1. Doesn't the compiler do this conversion for me? > > float x, y; > y = 0.5 * x; > > The conversion is effectively 'y = 0.5 * (double)x' where > now the rhs is evaluated in double (53-bit precision). If > you have 'y = 0.5f * x', then the rhs side is evaluate > in float (24-bit precision). For a more complicated, > expression whether one computes in 53 rather than 24 bits > can have an effect on the outcome. > Thanks for the clarification. I made the changes in catrigf.c adding "F"s as needed.
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