解释了Zsh完成缓存策略

Question

我正在使用强大的完成功能编写一些zsh函数。计算我的完成需要一些时间，我想使用完成缓存策略。从zsh手册(https://zsh.sourceforge.io/Doc/Release/Completion-System.html)中，我找到了这个代码片段

example_caching_policy () {
    # rebuild if cache is more than a week old
    local -a oldp
    oldp=( "$1"(Nm+7) )
    (( $#oldp ))
}

我找不到关于(Nm+7)语法的任何解释，Nm是什么意思？通过尝试和错误，我可以发现，例如，Nms+1将将缓存策略更改为1秒，而Nmh+1将更改为1小时。但是我在哪里可以找到一般的(NmX+N)构造解释呢？同样，这一行(( $#oldp ))到底是什么意思？

Answer 1

我可以解释一下(Nm+7)

man zshexpn，搜索Glob限定符

a[Mwhms][-|+]n
    files accessed exactly n days ago.  Files accessed within the last n days are selected using a negative
    value for n (-n).  Files accessed more than n days ago are selected by a positive n  value  (+n).   Op‐
    tional  unit  specifiers  `M',  `w',  `h', `m' or `s' (e.g. `ah5') cause the check to be performed with
    months (of 30 days), weeks, hours, minutes or seconds instead of days, respectively.  An  explicit  `d'
    for days is also allowed.

    Any  fractional  part of the difference between the access time and the current part in the appropriate
    units is ignored in the comparison.  For instance, `echo *(ah-5)' would echo files accessed within  the
    last  five  hours,  while  `echo  *(ah+5)'  would  echo files accessed at least six hours ago, as times
    strictly between five and six hours are treated as five hours.

m[Mwhms][-|+]n
    like the file access qualifier, except that it uses the file modification time.

N代表NULL_GLOB，如果zsh不匹配，它将删除模式。

如果没有这个N选项，如果不匹配，它将打印一个错误。

有4个文件的示例

$ touch lal # = updates file modification date to now
$ ls
lal  lil  lol  tut

$ ls l*(m+7)
lil  lol 
# files older than 7 days starting with l

$ ls l*(m-7)
lal 
# files younger than 7 days starting with l

$ ls l*(m+200)
zsh: no match
# no files older than 200 days

$ ls l*(Nm+200)
lal  lil  lol  tut
# N = NULL_GLOB made disappear the non-matching pattern so it's just ls