Date: Wed, 16 Jan 2013 10:14:19 -0800 From: "Thomas D. Dean" <tomdean@speakeasy.org> Cc: "questions@FreeBSD.org" <questions@FreeBSD.org> Subject: Re: time_t definition Message-ID: <50F6EDFB.70501@speakeasy.org> In-Reply-To: <20130116120015.3b8d0db4@mr129166> References: <50F5A189.7000701@speakeasy.org> <20130116120015.3b8d0db4@mr129166>
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On 01/16/13 03:00, Patrick Lamaiziere wrote:
> Looks like gcc47 checks the printf format string (-Wformat)
> Disable this check or convert your time_t.
Yes, I know gcc47 checks the format string.
But, time_t is of type int32, from a typedef statement.
#include <stdio.h>
typedef int zzz;
typedef zzz yyy;
typedef yyy xxx;
int main() {
xxx idx;
for (idx=0; idx<10; idx++) printf("%d\n",idx);
return 0;
}
does not produce the error (I did this on the 'other' system)
> gcc --version
gcc (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3
...
> gcc -O2 -pipe -I../../include -std=gnu99 -fstack-protector
-Wsystem-headers -Werror -Wall -Wno-format-y2k -Wno-uninitialized
-Wno-pointer-sign xxx.c -o xxx
I did not think to do this on the FreeBSD system I was using yesterday.
What I don't understand is where gcc is losing track of this definition.
In 9.0, or maybe earlier, the definition of time_t was changed with a
view toward 64-bit systems. I remember a statement to the effect of "in
2038, 32-bit time will overflow. It is unlikely that many 32-biot
systems will be around then. So, making the change to 64-bit now will
prevent having to do it in the future".
So, now, it seems that any calculation involving time_t requires a cast????
Tom Dean
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