Question

I am using an application which takes File X as input and gives File Y and Z as output (graphs and plots)
Now if the applicatoin is run again :

File Y and Z already exists and File X is not updated then it should not be over-written.
File X is updated, then File Y and Z must be overwritten.

法典在C++中写成,如何改变法典,从而满足这两种条件? 我是否应当每次拿X的邮票?

Answer 1

If you are on a POSIX machine, use stat or a related function such as fstat. The windows equivalents are described at http://msdn.microsoft.com/en-us/library/14h5k7ff(v=vs.80).aspx.

由stat归还的结构或窗口对口单位有一个代表文件最后修改时间的领域。

Answer 2

Answer 3

You could just keep a hash of file X. If the has changes, the file has been changed (updated), and you will need to update Y and Z.

if (XHashPrev != XHashNow || !YZExist)
    // update files.

Answer 4

如果X的修改晚于Y或Z,则使用David Hammen的回答,对X的修改时间长度与Y和Z进行比较,因此这些修改已经过时,应当改写。

Answer 5

是的,你应检查第十卷,看看(两种选择):

首先是最容易和最快的。第二是最准确的。 <>><>EDIT further to the comments below: 这两项备选办法可以结合起来,达成妥协。

Answer 6

另一种可能性是:用X书写基本方案,总是超过Y和Z字。加入基本方案。一份概述如下的文件:

Y Z : X
    basicProgram X Y Z

should do the trick. You can then invoke it with either "make Y" or "make Z" Bottom line: let make worry about timestamps and such.

Answer 7

...... 在某些档案系统中,[时间]的分辨率低到1小时。在方案执行期间,系统锁可能会被其他一些可能自动程序设定为新的价值......。