Date: Tue, 2 Mar 1999 12:03:04 -0500 (EST) From: "Crist J. Clark" <cjc@cc942873-a.ewndsr1.nj.home.com> To: freebsd-questions@FreeBSD.ORG (FreeBSD Questions) Subject: FP Math Message-ID: <199903021703.MAA02135@cc942873-a.ewndsr1.nj.home.com>
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I have been having some trouble with floating point
operations, namely the dreaded,
Floating exception (core dumped)
Having a look at 'man math' only really served to confuse me more. The
manpage lays out information about two libraries, DEC VAX-11
D_floating-point and IEEE STANDARD 754 Floating-Point
Arithmetic. However, no where does the manpage really come clean and
admit what we are using. I _think_ I am using the IEEE Standard, but
have been getting weird results. See below for an example. I think the
problems are related to float-double issues.
Why does the manpage inform me about both? Does it mean I can somehow
select one or the other? If I can, how do I do it?
Thanks.
Example: Here are two quick C program that I think show why I am
confused. The first produces an exception and core dumps (for me
anyway). In this one, I have set floats to values outside of their
range. It dumps at the point marked,
--- C Code ---
#include <stdio.h>
int main()
{
float a = -1.0e-41;
float b = -1.0e-41;
float c = 1.0e-41;
float d;
printf("%g\n%g\n",a,c); /* This outputs values a and b different
than those set */
d = a/b; /* This works */
printf("%g\n",d); /* Still OK */
d /= c; /* FP exception, core dump */
printf("%g\n",d);
return 0;
}
--- End C Code ---
Now the problem printing I could understand. I have set the floats
outside of their range. However, if I change the initial values
slightly,
--- C Code ---
#include <stdio.h>
int main()
{
float a = -1.0e41;
float b = -1.0e-41;
float c = 1.0e-41;
float d;
printf("%g\n%g\n",a,c); /* This gets a as it should be now */
d = a/b; /* This works */
printf("%g\n",d); /* Still OK */
d /= c; /* now this works! */
printf("%g\n",d);
return 0;
}
--- End C Code ---
Could someone explain this?
[Should I send this to -hackers instead?]
--
Crist J. Clark cjclark@home.com
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