如何获得Perl系统调用的CPU时间? [英] How can I get the CPU time for a perl system call?
问题描述
我有一个Perl脚本，该脚本使用system()调用外部可执行文件.我想测量这些外部程序占用的CPU秒数.理想情况下，我想使用shell内置time命令运行它们(在Linux系统上).像这样:
I have a perl script that calls external executables using system(). I would like to measure the CPU seconds taken by these external programs. Ideally, I would like to run them using the shell builtin time command (this is on a Linux system). Something like this:
system("time /path/to/command")
现在，time将其输出打印到stderr，但要这样做，请在单独的子shell中启动给出的命令.这意味着，为了在外壳程序中手动运行时捕获时间输出，您需要显式使用子外壳程序并重定向该子外壳程序的stderr:
Now, time prints its output to stderr, but in order to do so, launches the command it is given in a separate subshell. This means that in order to capture time's output when running manually in the shell, you need to explicitly use a subshell and redirect the subshell's stderr:
$ time ( command > command.log 2> command.er) 2> time.out
文件time.out将具有time命令的输出，而command.er具有stderr command.不幸的是，括号破坏了perl的系统调用:
The file time.out will have the output of the time command while command.er has the stderr of command. Unfortunately, the parentheses break perl's system call:
$ time ( ls ) 2> er ## works
$ perl -e 'system("time (ls)")'
sh: 1: Syntax error: word unexpected (expecting ")")
这意味着我无法捕获time的输出.更令人担忧的是，这似乎与版本有关:
And this means I can't capture the output of time. To make matters wors, this seems to be version dependent:
$ perl --version | head -n2
This is perl 5, version 18, subversion 2 (v5.18.2) built for x86_64-linux-gnu-thread-multi
但是，如果我尝试使用更新的版本进行同样的操作:
But if I try the same thing with a newer version:
$ perl --version | head -n2
This is perl 5, version 24, subversion 1 (v5.24.1) built for x86_64-linux-thread-multi
$ perl -e 'system("time (ls)")'
file1
real 0m0.002s
user 0m0.000s
sys 0m0.000s
不幸的是，我需要在生产机器上运行它，因此无法升级Perl.那么，如何在Perl 5.18中计时系统调用呢?我需要user和sys值，因此仅记录开始时间和结束时间将无济于事.如果有必要，我愿意使用专用模块，尽管我更喜欢让我使用外壳的time的技巧.
Unfortunately, I need this to run on a production machine so upgrading Perl is not an option. So, how can I time a system call in Perl 5.18? I need the user and sys values, so simply recording the start and end times won't help. I am willing to use a dedicated module if that's necessary although I would prefer a trick that lets me use the shell's time.
更新:事实证明，行为上的差异不是由于更新的perl版本引起的，而是因为我在/bin/sh为bash的Arch系统上测试了它，而其他命令正在运行/bin/sh为dash的Ubuntu系统，这是一个不支持子外壳括号的最小外壳.
UPDATE: it turns out the difference in behavior is not because of the newer perl version but, instead, it is because I tested it on an Arch system whose /bin/sh is bash while the other commands were being run on Ubuntu systems whose /bin/sh is dash, a minimal shell that doesn't support parentheses for subshells.
推荐答案
您可以使用捕获:: Tiny 捕获Perl中几乎所有东西的STDOUT和STDERR.
You can use Capture::Tiny to capture the STDOUT and STDERR of pretty much anything in Perl.
use Capture::Tiny 'capture';
my ($stdout, $stderr, $exit) = capture { system "time ls" };
print $stderr;
由于某种原因，输出在我的系统上缺少一些空格，但足够清晰以解析出您需要的内容.
For some reason the output is missing some whitespace on my system, but is clear enough to parse out what you need.
0.00user 0.00system 0:00.00elapsed 0%CPU (0avgtext+0avgdata 2272maxresident)k
0inputs+8outputs (0major+111minor)pagefaults 0swaps
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