Parameter expansion (variable expansion) and quotes within quotes

While working in a (Unix) shell, sometimes I need to double-quote single-quotes or vice versa, i.e. to write something like "foo 'bar baz' moo" or 'foo "bar baz" moo'. This may be because

• the resulting string will be parsed again (like with eval, with a shell run by watch, with a shell run by sshd on some remote server etc.) and these inner quotes are crucial to distinguish actual arguments during this second parsing;
• or I just need the original argument to include quotes for whatever other reason.

Some ambiguity appears when instead of bar baz I have a variable, say $variable. This is what POSIX says:

2.2.2 Single-Quotes

Enclosing characters in single-quotes ('') shall preserve the literal value of each character within the single-quotes. A single-quote cannot occur within single-quotes.

2.2.3 Double-Quotes

Enclosing characters in double-quotes ("") shall preserve the literal value of all characters within the double-quotes, with the exception of the characters backquote, <dollar-sign>, and <backslash>, as follows:

$
The shall retain its special meaning introducing parameter expansion […], a form of command substitution […], and arithmetic expansion […].

[…]

It's not exactly obvious which rule applies for $variable inside single-quotes inside double-quotes (or inside double-quotes inside single-quotes), therefore I'm asking:

1. Which rule applies to:

   • variable=123
     echo "'$variable'"
     

   • variable=123
     echo '"$variable"'
     

   ^{(I'm aware this can be easily answered just by trying, it would be a reasonable research effort. I decided to put this effort into the community wiki answer below).}

2. Are there any exceptions, quirks, surprises?

   ^{(Please note there is a community wiki answer below. Instead of adding a separate answer with another exception or quirk, consider contributing to the community wiki one).}

Solution 1:

Which rule applies?

Just by trying:

• $ variable=123
  $ echo "'$variable'"
  '123'
  $
  

• $ variable=123
  $ echo '"$variable"'
  "$variable"
  $
  

So the outer quotes matter. Common shells behave in the same way: sh, bash, dash, zsh, … It's because of this general rule:

If the current character is <backslash>, single-quote, or double-quote and it is not quoted, it shall affect quoting for subsequent characters up to the end of the quoted text.

^(source)

Only the outer quotes are not quoted, the rule applies to them. The inner quotes are quoted, they do not affect subsequent characters.

Are there any exceptions, quirks, surprises?

^{(Well, echo may introduce few surprises, see Why is printf better than echo? Still, this is a separate issue. For readability the answer uses echo with a safe value of $variable).}

1. The above code gives us either '123' or "$variable". What if we need "123" or '$variable'?

   • To get "123":

     variable=123
     echo "\"$variable\""
     #     ^^         ^^   escaped inner quotes
     

     or

     variable=123
     echo '"'"$variable"'"'
     #    1 12         23 3   outer quotes (numbered pairs)
     #     ^  ^^^^^^^^^  ^    quoted fragments
     

   • To get '$variable':

     variable=123
     echo "'"'$variable'"'"
     #    1 12         23 3   outer quotes (numbered pairs)
     #     ^  ^^^^^^^^^  ^    quoted fragments
     

2. The question notices that $ retains its special meaning in double-quotes. This is important:

   The input characters within the quoted string that are also enclosed between $( and the matching ) shall not be affected by the double-quotes, but rather shall define that command whose output replaces the $(…) when the word is expanded.

   ^(source)

   It means quoting within $(…) should be considered independently, despite the outer double-quotes. Indeed:

   $ variable=123
   $ echo "$(echo 'foo $variable')"
   $ #            ^             ^   single-quotes matter!
   foo $variable
   $
   

   (Note echo $(stuff) is bad code in general; it's here to illustrate the current issue).

   In these two cases:

   • echo "'$variable'" (tried way above)
   • echo "$(echo 'foo $variable')" (discussed here)

   there is $variable inside single-quotes inside double-quotes, but $(…) changes a lot. Similar situation with echo "`echo 'foo $variable'`" (although $(…) and backticks are not exactly equivalent).