wait for a time

I have a requirement to start my application at certain time. I don’t want to be put in the corn job. My executable is an application and like to start on 2011-Jan-20 So I have to run it as ./app –date 2011-Jan-20

Here problem is, how I will calculates the time difference from current and date supplied in command line 开发者_运维问答option.

I don’t want to write won function. Is there any in build function are available for such type of time difference. ( c and Linux)

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I know you're expecting a C answer but this might interest you:

Since you're on linux, the system provides already an efficient way to schedule ponctual tasks: at

In your case, an user that would like to run his task on the 20.01.2011 at 8AM, would just type:

echo "./app" | at 08:00 20.01.2011

The task will be run using the credentials of the user. Note that at also accept relative time directives such as at now +1 day. It is a powerful tool which ships with most Linux distributions by default.

The list of scheduled jobs can be get using:

atq

And you can even remove scheduled jobs using:

atrm

Hope this helps.

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You can calculate the difference between the start time and now in milliseconds and then wait for that many milliseconds by passing that number as a timeout argument to select() or epoll().

To calculate the difference, one way is to first convert your date string to struct tm using strptime() and then pass it to mktime() which is going to give you a number of seconds since unix epoch 1970-01-01 00:00:00. Then get the current time by using gettimeofday() or clock_gettime(), they also report time passed since unix epoch. Convert the start time and the current time to seconds and subtract the values.