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shellsedshbacktickscommand-substitution

double backslashes of sed single-quoted command inside command substitutions get translated to a single backslash

printf '%s' 'abc' | sed 's/./\\&/g'                        #1, \a\b\c
printf '%s' "`printf '%s' 'abc' | sed 's/./\\&/g'`"        #2, &&&

The expression inside the second backticks returns \a\b\c, and we have printf '%s' "\a\b\c", so it should print \a\b\c. My question is: why does the second script print &&& ?

note: I can get the second script work (prints \a\b\c) by prepending each backslash with another backslash, but I don't know why it's needed.

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

Solution

  • This is a good example to show difference between back-tick and $(cmd) command substitutions.

    When the old-style backquoted form of substitution is used, backslash retains its literal meaning except when followed by "$", "`", or "\". The first backticks not preceded by a backslash terminates the command substitution. When using the "$(COMMAND)" form, all characters between the parentheses make up the command; none are treated specially.

    http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html

    So take a look your example, I used echo instead of printf:

    kent$  echo 'abc' | sed 's/./\\&/g'
\a\b\c

kent$  echo -E "`echo 'abc' | sed 's/./\\&/g'`"
&&&

kent$  echo -E "$(echo 'abc' | sed 's/./\\&/g')"                    
\a\b\c

    You can see, the back-tick command substitution made your \\ as single \, thus together with the followed & it became \& (literal &)

    Note that I used echo -E in order to disable the interpretation of backslash escapes so that the \a\b\c could be printed out.