解释 Perl “序言"的狡猾之处 [英] Explain the deviousness of the Perl "preamble"

问题描述

Perl 手册描述了一个完全曲折的结构，可以在 csh、sh 或 Perl 中的任何一个下工作，例如以下:

eval '(exit $?0)' &&eval 'exec perl -wS $0 ${1+"$@"}'&eval 'exec/usr/bin/perl -wS $0 $argv:q'如果 $running_under_some_shell;

确实很狡猾……有人可以详细解释一下这是如何工作的吗?

解决方案

这个想法是，如果这三行在标准 Bourne shell (sh)、C shell (csh) 或珀尔.仅在不支持在脚本开头使用 #! 行指定解释器名称的系统上才需要此 hack.如果您执行以这 3 行开头的 Perl 脚本作为 shell 脚本，shell 将启动 Perl 解释器，将脚本的文件名和命令行参数传递给它.

在 Perl 中，三行构成一个语句，以 ; 结尾，形式为

eval '...' &&评估'...' &eval '...' 如果 $running_under_some_shell;

由于脚本刚刚启动，$running_under_some_shell 是 undef，这是错误的，并且永远不会执行 eval.这是一个空操作.

前面的部分是 $?0 在 sh 和 csh 中的解析方式不同.在sh中，这意味着$?(最后一条命令的退出状态)后跟0.由于没有上一条命令，$?将为0，所以$?0 计算结果为 00.在 csh 中，$?0 是一个特殊变量，如果当前输入文件名已知，则为 1，否则为 0.由于 shell 从脚本中读取这些行，$?0 将为 1.

因此，在 sh 中，eval '(exit $?0)' 的意思是 eval '(exit 00)'，而在 csh 中，它的意思是 eval '(出口1)'.括号表示应该在子 shell 中评估 exit 命令.

sh 和 csh 都理解 && 的意思是执行上一个命令，然后仅当上一个命令退出 0 时才执行以下命令".所以只有 sh 会执行 eval 'exec perl -wS $0 ${1+"$@"}'.csh 将进入下一行.

csh 将忽略行首的&".(我不确定这对 csh 到底意味着什么.从 Perl 的角度来看，它的目的是使它成为一个单一的表达式.)然后 csh 继续评估 eval 'exec/usr/bin/perl -wS $0 $argv:q'.

这两个命令行非常相似.exec perl 表示通过启动perl 的副本来替换当前进程.-wS 与 -w(启用警告)和 -S(在 $PATH 中查找指定脚本)的含义相同代码>).$0 是脚本的文件名.最后，${1+"$@"} 和 $argv:q 生成当前命令行参数的副本(分别在 sh 和 csh 中).>

它使用 ${1+"$@"} 而不是更常见的 "$@" 来解决一些古老版本的 Bourne shell 中的错误.他们的意思是一样的.您可以在 Bennett Todd 的解释(复制到 gbacon 的回答中)中阅读详细信息.

The Perl manual describes a totally devious construct that will work under any of csh, sh, or Perl, such as the following:

eval '(exit $?0)' && eval 'exec perl -wS $0 ${1+"$@"}'
    & eval 'exec /usr/bin/perl -wS $0 $argv:q'
    if $running_under_some_shell;

Devious indeed... can someone please explain in detail how this works?

解决方案

The idea is that those three lines do 3 different things if they're evaluated in a standard Bourne shell (sh), a C shell (csh), or Perl. This hack is only needed on systems that don't support specifying an interpreter name using a #! line at the start of a script. If you execute a Perl script beginning with those 3 lines as a shell script, the shell will launch the Perl interpreter, passing it the script's filename and the command line arguments.

In Perl, the three lines form one statement, terminated by the ;, of the form

eval '...' && eval '...' & eval '...' if $running_under_some_shell;

Since the script just started, $running_under_some_shell is undef, which is false, and the evals are never executed. It's a no-op.

The devious part is that $?0 is parsed differently in sh versus csh. In sh, that means $? (the exit status of the last command) followed by 0. Since there is no previous command, $? will be 0, so $?0 evaluates to 00. In csh, $?0 is a special variable that is 1 if the current input filename is known, or 0 if it isn't. Since the shell is reading these lines from a script, $?0 will be 1.

Therefore, in sh, eval '(exit $?0)' means eval '(exit 00)', and in csh it means eval '(exit 1)'. The parens indicate that the exit command should be evaluated in a subshell.

Both sh and csh understand && to mean "execute the previous command, then execute the following command only if the previous command exited 0". So only sh will execute eval 'exec perl -wS $0 ${1+"$@"}'. csh will proceed to the next line.

csh will ignore "& " at the beginning of a line. (I'm not sure exactly what that means to csh. Its purpose is to make this a single expression from Perl's point of view.) csh then proceeds to evaluate eval 'exec /usr/bin/perl -wS $0 $argv:q'.

These two command lines are quite similar. exec perl means to replace the current process by launching a copy of perl. -wS means the same as -w (enable warnings) and -S (look for the specified script in $PATH). $0 is the filename of the script. Finally both ${1+"$@"} and $argv:q produce a copy of the current command line arguments (in sh and csh, respectively).

It uses ${1+"$@"} instead of the more usual "$@" to work around a bug in some ancient version of the Bourne shell. They mean the same thing. You can read the details in Bennett Todd's explanation (copied in gbacon's answer).

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