如何在linux内核空间中获取当前小时（一天中的时间）
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How to get current hour (time of day) in linux kernel space
提问by Tom Macdonald
I'm writing a kernel module that checks to see if the time is between two specified hours, and disables input if it is. This has to do with me wanting to make sure I go to bed early. (I know I could also use any number of different techniques including cron etc, but I wanted to learn kernel programming...)
我正在编写一个内核模块，用于检查时间是否在两个指定的小时之间，如果是，则禁用输入。这与我想确保我早点睡觉有关。（我知道我也可以使用任何数量的不同技术，包括 cron 等，但我想学习内核编程......）
As a first version, I therefore check if the current hour is between start and end, which are set via parameters to the module.
作为第一个版本，我因此检查当前小时是否在开始和结束之间，这是通过模块的参数设置的。
My question is therefore : How do I get the current hour? I have no access to the usual time functions in the standard library because I am in kernel space. I'm guessing that I should be using do_gettimeofday() for this, but that only gives me seconds and nanoseconds, and I need hours in the current day.
因此，我的问题是：如何获取当前小时数？我无法访问标准库中的常用时间函数，因为我在内核空间中。我猜我应该为此使用 do_gettimeofday() ，但这只会给我几秒和几纳秒，而我在当天需要几个小时。
Thanks.
谢谢。
采纳答案by Zimbabao
time_to_tmfunction can be of your help, which returns the structure tm. Timezone available in variable sys_tz, it can help you to set your offset properly to get local time.
time_to_tm函数可以为您提供帮助，它返回结构tm。变量sys_tz 中可用的时区，它可以帮助您正确设置偏移量以获得本地时间。
回答by Erik
Converting the do_gettimeofday result to an hour is pretty simple, since it starts at midnight GMT.
将 do_gettimeofday 结果转换为一个小时非常简单，因为它从格林威治标准时间午夜开始。
time_t t = time(0);
time_t SecondsOfDay = t % (24*60*60);
time_t HourGMT = SecondsOfDay / (60*60);
Then adjust for your local timezone
然后根据您当地的时区进行调整
回答by sunmoon
We can use clock_gettime function with CLOCK_REALTIME as the type of clock.
我们可以使用clock_gettime 函数和CLOCK_REALTIME 作为时钟类型。
Reference http://linux.die.net/man/3/clock_gettime
参考http://linux.die.net/man/3/clock_gettime
Just doing a strace on date executable gives us an idea to get the current date in the kernel mode.
只需在日期可执行文件上执行 strace 就可以让我们了解在内核模式下获取当前日期。
回答by Roman
This works well for me:
这对我很有效：
#include <linux/time.h>
...
/* getnstimeofday - Returns the time of day in a timespec */
void getnstimeofday(struct timespec *ts)
For getting usual time format you can use:
要获得通常的时间格式，您可以使用：
printk("TIME: %.2lu:%.2lu:%.2lu:%.6lu \r\n",
(curr_tm.tv_sec / 3600) % (24),
(curr_tm.tv_sec / 60) % (60),
curr_tm.tv_sec % 60,
curr_tm.tv_nsec / 1000);
回答by Bharath
To get the local time in kernel, add the below code snippet your kernel driver:
要获取内核中的本地时间，请在内核驱动程序中添加以下代码片段：
struct timeval time;
unsigned long local_time;
do_gettimeofday(&time);
local_time = (u32)(time.tv_sec - (sys_tz.tz_minuteswest * 60));
rtc_time_to_tm(local_time, &tm);
printk(" @ (%04d-%02d-%02d %02d:%02d:%02d)\n", tm.tm_year + 1900, tm.tm_mon + 1, tm.tm_mday, tm.tm_hour, tm.tm_min, tm.tm_sec);
