为什么要引用$？明确地

发布于 2024-08-16 05:01:34 字数 211 浏览 4 评论 0原文

很多 sh 代码看起来像：

$cmd
if [ $? = 0 ]; then $cmd2; fi

而不是：

if $cmd; then $cmd2; fi

我通常假设人们使用前者只是因为他们不知道后者的语法是有效的，但我想知道是否还有其他原因（尽管唯一的可能性是我想到的是便携性）。是否有任何理由更喜欢显式引用 ${?}?。

原文

A lot of sh code looks like:

$cmd
if [ $? = 0 ]; then $cmd2; fi

Instead of:

if $cmd; then $cmd2; fi

I have typically assumed that people use the former simply because they are unaware that the syntax of the latter is valid, but I'm wondering if there is another reason (although the only possibility that comes to mind is portability). Is there any reason to prefer explicitly referencing ${?}?.

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深居我梦 2024-08-23 05:01:34

如果您确实想将错误代码用于某些用途（日志记录、处理预期错误等），则必须引用 $?。对于您提到的特定场景，第二个 if 构造更清晰、更短，我认为没有理由不使用它。

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梦里南柯 2024-08-23 05:01:34

在我看来，因为很多人没有意识到你可以或多或少地测试命令的状态。同样，人们也没有意识到你可以写：

if cmd1
   cmd2
   cmd3
then
    ...do this after executing cmd1, cmd2 and cmd3, but ...
    ...only if cmd3 exits with with status 0!
fi

你还可以将你的代码简化（以简洁为代价）为：（

cmd1 && cmd2

我看到@shin在评论中指出了这一点，我已经投了赞成票）。

我还看到很多“Bourne”shell 代码使用：

if ( ... )
then ...
fi

有时这是合适的 - 但大多数情况下，它是 C shell 程序员编写的东西，没有意识到该符号在 Bourne shell 和衍生产品中意味着“运行子 shell” -例如 Korn shell、符合 POSIX 标准的 shell 和 Bash。

In my view, because a lot of people don't realize that you can test the status of the command more or less as you do. Similarly, people also do not realize that you can write:

if cmd1
   cmd2
   cmd3
then
    ...do this after executing cmd1, cmd2 and cmd3, but ...
    ...only if cmd3 exits with with status 0!
fi

You can also simplify your code (at the cost of terseness) to:

cmd1 && cmd2

(I see @shin noted this in a comment, which I've up-voted).

I also see a lot of 'Bourne' shell code that uses:

if ( ... )
then ...
fi

Sometimes that is appropriate - but most often it is stuff that a C shell programmer wrote not realizing that the notation means 'run a sub-shell' in Bourne shell and derivatives - such as Korn shell, the POSIX-compliant shells, and Bash.

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