From: "Kaveh R. GHAZI" <ghazi@caip.rutgers.edu>
To: gcc@gcc.gnu.org
Subject: What is -3.I (as opposed to 0-3.I) supposed evaluate to?
Date: Mon, 08 Jun 2009 20:12:00 -0000 [thread overview]
Message-ID: <Pine.GSO.4.58.0906081603270.4648@caipclassic.rutgers.edu> (raw)
If I write a complex double constant -3.I (as opposed to 0-3.I), what is
it supposed to evaluate to? This program:
#include <stdio.h>
int main(void)
{
const __complex double C1 = (-3.I);
const __complex double C2 = (0-3.I);
printf ("%f %f\n", __real__ C1, __imag__ (C1));
printf ("%f %f\n", __real__ C2, __imag__ (C2));
return 0;
}
when compiled with gcc-4.1.2 (and mainline) yields:
-0.000000 -3.000000
0.000000 -3.000000
Note the sign difference in the real part.
When I compile it with g++-4.1.2, I get:
compl.c: In function 'int main()':
compl.c:5: error: wrong type argument to unary minus
Is this supposed to happen or is it a bug in complex number parsing?
(Sorry if this is a gcc-help question.)
Thanks,
--Kaveh
next reply other threads:[~2009-06-08 20:12 UTC|newest]
Thread overview: 6+ messages / expand[flat|nested] mbox.gz Atom feed top
2009-06-08 20:12 Kaveh R. GHAZI [this message]
2009-06-08 20:33 ` Joseph S. Myers
2009-06-09 2:08 ` Kaveh R. Ghazi
2009-06-09 11:22 ` Joseph S. Myers
2009-06-09 15:26 ` Kaveh R. Ghazi
2009-06-09 15:33 ` Richard Guenther
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