perl - 当我使用非零参数显式调用 exit 时，为什么我的 Perl 脚本会返回零返回码？

Question

我有一个调用另一个脚本的 Perl 脚本。 Perl 脚本应该传播脚本的返回代码，但似乎向其调用者(Java 应用程序)返回零，尽管显式调用 exit $scriptReturnCode .

代码和输出如下(我意识到 <=> 可以/应该是 != 但这就是我所拥有的):

print "INFO: Calling ${scriptDirectory}/${script} ${args}"
$scriptReturnCode = system("${scriptDirectory}/${script} ${args}");

if ( $scriptReturnCode <=> 0 ) {
        print "ERROR: The script returned $scriptReturnCode\n";
        exit $scriptReturnCode;
} else {
        print "INFO: The script returned $scriptReturnCode.\n";
        exit 0;
}

我的 Java 输出是:

20/04/2010 14:40:01 - INFO: Calling /path/to/script/script.ksh arg1 arg2 
20/04/2010 14:40:01 - Could not find installer files <= this is from the script.ksh    
20/04/2010 14:40:01 - ERROR: The script returned 256
20/04/2010 14:40:01 - Command Finished. Exit Code: 0 <= this is the Java app.

Answer 1

您需要将返回码从 system() 移开按 8 位调用。

例如。 $exit_value = $? >> 8; # 在你的脚本中 $?是 $scriptReturnCode

来自 http://perldoc.perl.org/perlfaq8.html :

system() runs a command and returns exit status information (as a 16 bit value: the low 7 bits are the signal the process died from, if any, and the high 8 bits are the actual exit value

对核心转储的更扩展的代码检查也可能如下所示:

system();
if ($? == -1) {
    print "failed to execute: $!\n";
} elsif ($? & 127) {
    printf "child died - signal %d, %s coredump\n",
           ($? & 127), ($? & 128) ? 'with' : 'without';
} else {
    printf "child exited with value %d\n", $? >> 8;
}

更新:根据 ysth 的精彩提示，退出代码被截断为 8(低)位，因此返回 256 而不是预期的 1 最终为 0。同样，返回 257 最终为 1。