为什么这个单引号字符串在命令替换中被解释 [英] why does this single quoted string get interpreted when it's inside of a command substitution

问题描述

比较这两行shell脚本:

Compare these two lines of shell script:

printf '%s' 's/./\\&/g'                    #1, s/./\\&/g
printf '%s' `printf '%s' 's/./\\&/g'`      #2, s/./\&/g

我的问题是:为什么单引号双反斜杠被解释为第二行脚本的单反斜杠?

My question is: why does the single-quoted double backslashes get interpreted as a single backslash for the second line of script?

推荐答案

从

printf '%s' `printf '%s' 's/./\\&/g'`

反引号内的表达式返回 s/./\\&/g 与第一个表达式一样，没有单引号，所以你得到

The expression inside backticks returns s/./\\&/g as in the first expression, without single quotes, so you get

printf '%s' s/./\\&/g

第一个反斜杠将第二个反斜杠转义，因此打印s/./\&/g.

The first backslash escapes the second one, so it prints s/./\&/g.

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