Why does strptime() behave differently on OSX and on Linux?

问题

Consider this program:

#include <stdio.h>
#include <time.h>

int main() {
  struct tm t;
  strptime("2015-08-13 12:00:00", "%F %T", &t);
  printf("t.tm_wday = %d\n", t.tm_wday);
  return 0;
}

Under OSX, this is what I obtain:

$ gcc test_strptime.c
$ ./a.out
t.tm_wday = 0

But on Linux, this is what I get:

$ gcc test_strptime.c
$ ./a.out
t.tm_wday = 4

Why is the bahaviour different? I would expect the day of the week to be well defined, given the data and the time of the day?

回答1:

The Linux (glibc) and OS X implementations of strptime are different. From the OS X man page:

If the format string does not contain enough conversion specifications to completely specify the resulting struct tm, the unspecified members of tm are left untouched. For example, if format is ``%H:%M:%S'', only tm_hour, tm_sec and tm_min will be modified.

Compared with glibc:

The glibc implementation does not touch those fields which are not explicitly specified, except that it recomputes the tm_wday and tm_yday field if any of the year, month, or day elements changed.

So, you could say it's working as documented.

来源：https://stackoverflow.com/questions/31984002/why-does-strptime-behave-differently-on-osx-and-on-linux