Question

我试图在perl中捕获一个python exit coode形式。

我的python代码py_test.py：

import sys
sys.exit(sys.argv[1])

我的perl代码test.pl：

#!/usr/bin/perl -w
$res = system("python py_test.py ".$ARGV[0]);
print("\$res ==== $res \n >>>". $? ."\n");

我跑了

perl test.pl 4

结果是：

4
$res ==== 256
>>>256

我在32位RHEL 5.3上使用python 2.4.3和perl 5.8.8。

有人可以帮助我了解这里发生的事情吗，我似乎无法理解这种行为，我做错了什么吗？（我看了几个小时的代码，从未遇到过系统怪异的东西）

Answer 1

我发现了问题。我输入sys.exit argv [1]，但是在开始时python将其视为字符串。将其更改为sys.exit（int（argv [1]））后，一切正常。

这也解释了为什么打印“ 4” ...

Answer 2

来自“ perldoc perlvar”

$?      The status returned by the last pipe close, backtick (``) command,
        successful call to "wait()" or "waitpid()", or from the "system()"
        operator. This is just the 16-bit status word returned by the
        traditional Unix "wait()" system call (or else is made up to look
        like it). Thus, the exit value of the subprocess is really ("$? >>
        8"), and "$? & 127" gives which signal, if any, the process died
        from, and "$? & 128" reports whether there was a core dump.

听起来您的python脚本可能无法成功退出4。

test.py

#!/usr/bin/env python3

import sys

sys.exit(4)

test.pl

#!/usr/bin/env perl
use warnings;
use strict;

system("./test.py");
print "\$? = $?\n"
die("Killed by signal ".( $? & 0x7F ) if $? & 0x7F;
die("Exited with error ".( $? >> 8 ) if $? >> 8;
print "Success\n";

结果：

~/test$ ./test.pl
$? = 1024
Exited with error 4

从Perl调用Python脚本

2 个答案: