CODEWITHSUNDEEP
bashshellparameter-expansion
mickp
mickp

Reputation: 1809

${foo//(/\\(} not working with extglobs enabled

I am trying to escape parentheses using parameter expansion. Although if I have extglob enabled, the following code doesn't work:

#!/usr/bin/env bash

shopt -s extglob

foo='file(2)'
foo=${foo//(/\\(}
foo=${foo//)/\\)}

printf '%s\n' "$foo"

# Expected:  file\(2\)
# Actual:    file(2\)

It correctly outputs file\(2\) when I disable extglob or explicitly escape the left parenthesis like this:

foo=${foo//\(/\\(}

Why does extglob cause that? I don't see any extglob pattern there. Also, the right parenthesis works fine without a backslash.

Tested online at tutorialspoint.com and also locally using:

GNU bash, version 4.3.30(1)-release (x86_64-unknown-linux-gnu)
GNU bash, version 4.4.18(1)-release (x86_64-unknown-linux-gnu)
GNU bash, version 5.0.0(2)-alpha (x86_64-pc-linux-gnu)

Upvotes: 3

Views: 100

Answers (2)

that other guy
that other guy

Reputation: 123490

This is a bug due to an optimization in bash.

When replacing a pattern, bash first checks whether the pattern matches anywhere in the string. If it doesn't, then there's no point in doing any search&replace. The way it does it is by construct a new pattern by surrounding it with *..* as necessary:

  /* If the pattern doesn't match anywhere in the string, go ahead and
     short-circuit right away.  A minor optimization, saves a bunch of
     unnecessary calls to strmatch (up to N calls for a string of N
     characters) if the match is unsuccessful.  To preserve the semantics
     of the substring matches below, we make sure that the pattern has
     `*' as first and last character, making a new pattern if necessary. */
  /* XXX - check this later if I ever implement `**' with special meaning,
     since this will potentially result in `**' at the beginning or end */
  len = STRLEN (pat);
  if (pat[0] != '*' || (pat[0] == '*' && pat[1] == LPAREN && extended_glob) || pat[len - 1] != '*')
    {
      int unescaped_backslash;
      char *pp;

      p = npat = (char *)xmalloc (len + 3);
      p1 = pat;
      if (*p1 != '*' || (*p1 == '*' && p1[1] == LPAREN && extended_glob))
    *p++ = '*';

The pattern it tries to match against the string ends up being *(*

The opening *( is now unintentionally recognized as the start of an extglob, but when bash fails to find the closing ), it matches the pattern as a string instead:

 prest = PATSCAN (p + (*p == L('(')), pe, 0); /* ) */
  if (prest == 0)
    /* If PREST is 0, we failed to scan a valid pattern.  In this
       case, we just want to compare the two as strings. */
    return (STRCOMPARE (p - 1, pe, s, se));

This means that unless the string to do replacements in is literally *(*, the optimization invalidly rejects the string thinking there's nothing to do. Of course, this also means that it works correctly for *(* itself:

$ f='*(*'; echo "${f//(/\\(}"
*\(*

If you were to fudge this optimization check in the source code:

diff --git a/subst.c b/subst.c
index fc00cab0..f063f784 100644
--- a/subst.c
+++ b/subst.c
@@ -4517,8 +4517,6 @@ match_upattern (string, pat, mtype, sp, ep)
   c = strmatch (npat, string, FNMATCH_EXTFLAG | FNMATCH_IGNCASE);
   if (npat != pat)
     free (npat);
-  if (c == FNM_NOMATCH)
-    return (0);

   len = STRLEN (string);
   end = string + len;

then it would work correctly in your case:

$ ./bash -c 'f="my string(1) with (parens)"; echo "${f//(/\\(}"'
my string\(1) with \(parens)

Upvotes: 3

Charles Duffy
Charles Duffy

Reputation: 295490

Quoting the search string prevents it from being interpreted as a glob, and thus moots the issue:

shopt -s extglob
foo='file(2)'
foo=${foo//'('/'\('}
foo=${foo//')'/'\)'}
printf '%s\n' "$foo"

(Quoting the replacement as well avoids the need to double up the backslashes).

Upvotes: 2

Related Questions