On 10/29/2010 6:16 PM, Eric Blake wrote:
On 10/29/2010 04:11 PM, Ken Brown wrote:
Thanks, Eric. I didn't know about any of this. (I was using a modification of
a configure test from the emacs sources.)
Probably worth pointing it out to the emacs upstream, then :)
But I get the same behavior with the following revised test case:
#include<time.h>
#include<stdio.h>
int
main (void)
{
time_t now = time ((time_t *) 0);
printf ("TZ is initially unset; hour = %d\n", localtime (&now)->tm_hour);
putenv ("TZ=GMT0");
printf ("TZ=GMT0; hour = %d\n", localtime (&now)->tm_hour);
unsetenv("TZ");
printf ("TZ unset; hour = %d\n", localtime (&now)->tm_hour);
putenv ("TZ=PST8");
printf ("TZ=PST8; hour = %d\n", localtime (&now)->tm_hour);
unsetenv("TZ");
printf ("TZ unset again; hour = %d\n", localtime (&now)->tm_hour);
}
So the question remains whether this difference between Cygwin and Linux is a
bug or by design.
Apparently by design. POSIX requires:
http://www.opengroup.org/onlinepubs/9699919799/functions/localtime.html
Local timezone information is used as though localtime() calls tzset().
http://www.opengroup.org/onlinepubs/9699919799/functions/tzset.html
The tzset() function shall use the value of the environment variable TZ
to set time conversion information used by ctime , localtime , mktime ,
and strftime . If TZ is absent from the environment,
implementation-defined default timezone information shall be used.
Wouldn't you interpret this as meaning that the implementation-defined
default timezone information should be the same every time localtime is
called with TZ unset? If not, what should a program do to get the
"standard" default timezone information that it would get if TZ had
never been set in the first place?
Ken
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