why does this single quoted string get interpreted when it's inside of a command substitution

Question

Compare these two lines of shell script:

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

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

score 2 · Accepted Answer · edited Jun 13 '16 at 15:53

2

Starting from

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

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

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

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

edited Jun 13 '16 at 15:53

Toby Speight

answered Jun 13 '16 at 12:01

WilQu

May I ask you one more related question, why does this prints &&&, not abc ? echo "\`echo 'abc' | sed 's/./\\&/g'\`" – pdg Jun 13 '16 at 12:28
@pdg you should ask another question. – WilQu Jun 13 '16 at 13:00
Ok, I'll wait and ask it in another hour. Thanks anyway. – pdg Jun 13 '16 at 13:12

1 Answers1