eval/bash -c 与仅评估变量相比有何意义? [英] What's the point of eval/bash -c as opposed to just evaluating a variable?

问题描述

假设您将以下命令存储在变量中:

Suppose you have the following command stored in a variable:

COMMAND='echo hello'

有什么区别

$ eval "$COMMAND"
hello

$ bash -c "$COMMAND"
hello

$ $COMMAND
hello

?如果最后一个版本更短并且(据我所知)做了完全相同的事情，为什么几乎从未使用过它?

? Why is the last version almost never used if it is shorter and (as far as I can see) does exactly the same thing?

推荐答案

第三种形式与其他两种形式完全不同——但要理解为什么，我们需要深入了解 bash 在解释命令时的操作顺序，并查看在使用每种方法时遵循了哪些方法.

The third form is not at all like the other two -- but to understand why, we need to go into the order of operations when bash in interpreting a command, and look at which of those are followed when each method is in use.

1. 报价处理
2. 拆分成命令
3. 特殊运算符解析
4. 扩展
5. 分词
6. 球状
7. 执行

<小时>

使用 eval "$string"

eval "$string" 从 #1 开始遵循上述所有步骤.因此:

---

Using eval "$string"

eval "$string" follows all the above steps starting from #1. Thus:

• 字符串中的文字引号变成句法引号
• 处理>()等特殊操作符
• 像$foo这样的扩展很受尊重
• 这些扩展的结果在字符上被拆分为空格并分成单独的单词
• 如果这些词解析为相同并且有可用的匹配项，则这些词会扩展为 globs，最后执行命令.

• Literal quotes within the string become syntactic quotes
• Special operators such as >() are processed
• Expansions such as $foo are honored
• Results of those expansions are split on characters into whitespace into separate words
• Those words are expanded as globs if they parse as same and have available matches, and finally the command is executed.

...执行与 eval 相同的操作，但在作为单独进程启动的新 shell 中；因此，当这个新进程退出时，对变量状态、当前目录等的更改将过期.(还要注意，新的 shell 可能是支持不同语言的不同解释器；即 sh -c "foo" 将不支持与 bash 相同的语法，ksh、zsh 等等.

...performs the same as eval does, but in a new shell launched as a separate process; thus, changes to variable state, current directory, etc. will expire when this new process exits. (Note, too, that that new shell may be a different interpreter supporting a different language; ie. sh -c "foo" will not support the same syntax that bash, ksh, zsh, etc. do).

...从第 5 步分词"开始.

...starts at step 5, "Word Splitting".

这是什么意思?

printf '%s ' "two words" 因此将解析为 printf %s "two words"，与 printf %s 两个字 的通常/预期行为相反(shell 使用引号).

printf '%s ' "two words" will thus parse as printf %s "two words", as opposed to the usual/expected behavior of printf %s two words (with the quotes being consumed by the shell).

因此:

s='echo foo && echo bar'
$s

...将发出以下输出:

...will emit the following output:

foo && echo bar

...而不是以下内容，否则会出现这种情况:

...instead of the following, which would otherwise be expected:

foo
bar

不支持特殊运算符和扩展.

没有$(foo)，没有$foo，没有<(foo)等

>foo 或 2>&1 只是由字符串拆分创建的另一个词，而不是 shell 指令.

>foo or 2>&1 is just another word created by string-splitting, rather than a shell directive.

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