如何对现有bash变量进行ANSI C引用? [英] How can I do ANSI C quoting of an existing bash variable?

问题描述

我查看了这个问题，但没有涵盖我的用例.

I have looked at this question, but it does not cover my use case.

假设我有一个变量 foo ，它包含四个字符的文字 \ x60 .

Suppose I have the variable foo which holds the four-character literal \x60.

我要执行 ANSI C报价在此变量的内容上，并将其存储到另一个变量 bar 中.

I want to perform ANSI C Quoting on the contents of this variable and store it into another variable bar.

我尝试了以下方法，但是都没有达到预期的效果.

I tried the following, but none of them achieved the desired effect.

bar=$'$foo'   
echo $bar     
bar=$"$foo"     
echo $bar       

输出:

$foo
\x61

所需的输出( \ x61 的实际值):

Desired output (actual value of \x61):

a

在一般情况下(包括不可打印的字符)，我该如何实现?请注意，在这种情况下，仅以 a 作为示例，以便更轻松地测试该方法是否有效.

How might I achieve this in the general case, including non-printable characters? Note that in this case a was used just as an example to make it easier to test whether the method worked.

推荐答案

到目前为止，如果您使用的是 bash :

By far the simplest solution, if you are using bash:

printf %b "$foo"

或者，将其保存到另一个变量名 bar :

Or, to save it in another variable name bar:

printf -v bar %b "$foo"

从 help printf :

除了printf(1)中描述的标准格式规范外和printf(3)，printf解释为:

In addition to the standard format specifications described in printf(1) and printf(3), printf interprets:

 %b        expand backslash escape sequences in the corresponding argument
 %q        quote the argument in a way that can be reused as shell input
 %(fmt)T output the date-time string resulting from using FMT as a format
         string for strftime(3)

尽管有一些极端情况:

\ c终止输出，不会删除\'，\和\?中的反斜杠，并且以\ 0开头的八进制转义字符最多可以包含四个数字

\c terminates output, backslashes in \', \", and \? are not removed, and octal escapes beginning with \0 may contain up to four digits

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