Why does the command echo `echo \\\\\\\z` in bash script print \\z instead of \\\z?

Question

The command

echo `echo \\\\\\\z`

is from this book , I don’t understand why it prints

\\z

when it get executed via the bash script.

I think it should print

\\\z

Answer 1

Backticks effectively give you double processing of the backslashes. (Once when the main command line is processed, and another time when the command in the command substitution is.)

Like Kusalanda says, that's different for the $(..) command substitution.

First, let's note that we appear to be talking about Bash's echo, which by default doesn't process backslash escapes itself, so e.g. echo '\\' passes \\ to echo which prints \\. If you used an echo that deals with backslashes itself (like Bash with shopt -s xpg_echo, or Dash, or zsh), you'd get that last \\ turned into \ and \z as the output.

---

Bash's manual says:

When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by [a dollar sign, backtick or backslash].

So, in the command

echo `echo \\\\\\\z`      # 7 backslashes

the \\ pairs get first reduced to single \s, and the \z stays as \z (it's not followed by any of the three, so that backslash is literal). Then the command in the backticks runs through the shell again.

So the command in the command substitution ends up as

echo \\\\z                 # 4 backslashes

where the \\ pairs are reduced to single \s, and the argument to echo becomes \\z (which is printed unchanged through the two echos, see caveat above).

You can see a similar result with a dollar sign, e.g.:

var=foo
echo `echo \$var`

prints foo.

POSIX has also a weird example with single quotes:

echo `echo '\$x'`

prints $x. (the single quotes don't protect the $)

---

The $(...) form of command substitution is saner, its contents only get processed once, so:

echo $(echo \\\\\\\z)           # 7 backslashes

prints \\\z (3), and

var=foo
echo $(echo \$var)

prints $var.

Also see:

• BashFAQ 082 "Why is $(...) preferred over ... (backticks)?"
• Have backticks (i.e. `cmd`) in *sh shells been deprecated?

Answer 2

The initial echo command is handled by the shell, and so it correctly sees the multiple \ and reduces them to echo \\\z.

At this point, the shell is no longer involved.

The echo command then sees \\\, and so it it reduces that to the single \, ie, you want a literal \.

There is no need to escape the last \, and so that then reduces to \\\z.