为什么strptime()在OSX和Linux上的行为不同? [英] Why does strptime() behave differently on OSX and on Linux?

问题描述

考虑此程序:

#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;
}

在OSX下，这是我得到的:

Under OSX, this is what I obtain:

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

但是在Linux上，这就是我得到的:

But on Linux, this is what I get:

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

为什么行为不同?考虑到数据和一天中的时间，我希望星期几能够得到很好的定义?

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?

推荐答案

strptime 的Linux(glibc)和OS X实现是不同的.从 OS X手册页:

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

如果格式字符串包含的转换规范不足以完全指定生成的 struct tm ，则tm的未指定成员保持不变.例如，如果格式为``％H:％M:％S''，只会修改 tm_hour ， tm_sec 和 tm_min .

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.

与 glibc 相比:

glibc实现不涉及那些未明确指定的字段，只是它会重新计算 tm_wday 和 tm_yday 字段(如果是年，月或日中的任何一个)元素已更改.

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.

因此，您可以说它按文档所述工作.

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

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