Bash – Why Does Variable Expansion Retain Quotes?

When instructed to echo commands as they are executed ("execution trace"), both bash and ksh add single quotes around any word with meta-characters (*, ?, ;, etc.) in it.

The meta-characters could have gotten into the word in a variety of ways. The word (or part of it) could have been quoted with single or double quotes, the characters could have been escaped with a \, or they remained as the result of a failed filename matching attempt. In all cases, the execution trace will contain single-quoted words, for example:

$ set -x
$ echo foo\;bar
+ echo 'foo;bar'

This is just an artifact of the way the shells implement the execution trace; it doesn't alter the way the arguments are ultimately passed to the command. The quotes are added, printed, and discarded. Here is the relevant part of the bash source code, print_cmd.c:

/* A function to print the words of a simple command when set -x is on. */
void
xtrace_print_word_list (list, xtflags)
...
{
  ...
  for (w = list; w; w = w->next)
    {
      t = w->word->word;
      ...
      else if (sh_contains_shell_metas (t))
        {
          x = sh_single_quote (t);
          fprintf (xtrace_fp, "%s%s", x, w->next ? " " : "");
          free (x);
        }

As to why the authors chose to do this, the code there doesn't say. But here's some similar code in variables.c, and it comes with a comment:

/* Print the value cell of VAR, a shell variable.  Do not print
   the name, nor leading/trailing newline.  If QUOTE is non-zero,
   and the value contains shell metacharacters, quote the value
   in such a way that it can be read back in. */
void
print_var_value (var, quote)
...
{
  ...
  else if (quote && sh_contains_shell_metas (value_cell (var)))
    {
      t = sh_single_quote (value_cell (var));
      printf ("%s", t);
      free (t);
    }

So possibly it's done so that it's easier to copy the command lines from the output of the execution trace and run them again.

The ' single quote character in your echo example gets it literal value (and loses its meaning) as it enclosed in double quotes ("). The enclosing characters are the double quotes.

What you can do is print the single quotes separately:

echo "'"'$a'"'"

or escape the $:

echo "'\$a'"