kayprish/timeoutd
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Use a temp to convert ut_time to time_t

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The localtime function takes a time_t* argument.  Passing
the ut_time member of the utmp struct as this argument
doesn't work on 64 bit systems because the lengths of the
arguments don't agree.  So use a temproary variable to
store the ut_time value in a time_t variable and pass
the address of the time_t variable to localtime.

Signed-off-by: Clark Rawlins <clark@bit63.org>
This commit is contained in:
Clark Rawlins 2014-12-14 10:29:17 -08:00
parent 4f37273c60
commit af2d3b48c6
1 changed files with 8 additions and 1 deletions
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9
timeoutd.c
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@ -398,6 +398,7 @@ void read_wtmp()
FILE *fp; FILE *fp;
struct utmp ut; struct utmp ut;
struct tm *tm; struct tm *tm;
time_t time;
#ifdef DEBUG #ifdef DEBUG
openlog("timeoutd", OPENLOG_FLAGS, LOG_DAEMON); openlog("timeoutd", OPENLOG_FLAGS, LOG_DAEMON);
@ -418,7 +419,13 @@ void read_wtmp()
while (fread(&ut, sizeof(struct utmp), 1, fp) == 1) while (fread(&ut, sizeof(struct utmp), 1, fp) == 1)
{ {
tm = localtime(&ut.ut_time); /* On 64 bit systems time_t is a 64 bit integer
while ut_time is a 32 bit (unsigned) integer.
Copy the value to a time_t value so the pointer
types agree in the call to localtime.
*/
time = ut.ut_time;
tm = localtime(&time);
if (tm->tm_year != now.tm_year || tm->tm_yday != now.tm_yday) if (tm->tm_year != now.tm_year || tm->tm_yday != now.tm_yday)
break; break;