mjhjmtu

Is there any consistent logic to it?

some-command "$somevariable//some pattern/'how does this get parsed?'"

I've posted some conclusions and raw tests below as an "answer" but they're not a full answer by any means. The Bash man page appears silent on the subject.

As discussed in the comments, this seems to have changed between versions of Bash. I think this is the relevant change in bash-4.3-alpha (changelog):

zz. When using the pattern substitution word expansion, bash now runs the
replacement string through quote removal, since it allows quotes in that
string to act as escape characters. This is not backwards compatible, so
it can be disabled by setting the bash compatibility mode to 4.2.

And the description for shopt -s compat42 (online manual):

compat42

If set, bash does not process the replacement string in the pattern substitution word expansion using quote removal.

The quoting single-quotes example:

$ s=abc'def; echo "'$s//'/'\'''"
'abc'''def'

$ shopt -s compat42
$ s=abc'def; echo "'$s//'/'\'''"
'abc'\''def'

$ bash --version | head -1
GNU bash, version 4.4.12(1)-release (x86_64-pc-linux-gnu)

Workaround: put the replacement string in a variable, and don't use quotes inside the replacement:

$ shopt -s compat42
$ qq="'''"; s=abc'def; echo "'$s//'/$qq'";
'abc'''def'
$ qq="'''"; s=abc'def; echo "'$s//'/"$qq"'";
'abc"'''"def'

The funny thing is, that if the expansion is unquoted, then the quotes are removed after the substitution, in all versions. That is s=abc; echo $s/b/"" prints ac. This of course doesn't happen with other expansions, e.g. s='a""c' ; echo $s%x outputs a""c.

General rules by reverse engineering:

• Quotes must be coupled (completed)


• Quotes are preserved (included in the actual replacement)


• Backslashes are preserved if they come before an arbitrary letter


• Backslashes are preserved if they escape a single quote


• A backslash backslash sequence is reduced to one backslash even within single quotes


• You can't escape a single quote within single quotes


• Parameter expansion works inside single quotes the same as outside


• If a dollar sign is escaped with a backslash the dollar sign is preserved literally and the backslash is removed

And a conclusion:

• There is absolutely no way to produce the literal sequence ''' as a substitution through parameter expansion.


• However, it is very easy to produce the literal sequence "'''" as a substitution.

Some raw tests follow.

[vagrant@localhost ~]$ echo "$0"
-bash
[vagrant@localhost ~]$ echo "$0//a/x"
-bxsh
[vagrant@localhost ~]$ echo "$0//a/some long string with spaces"
-bsome long string with spacessh
[vagrant@localhost ~]$ echo "$0//a/"quoted string""
-b"quoted string"sh
[vagrant@localhost ~]$ echo "$0//a/"unfinished quote"
> wat
> }"
-b"unfinished quote}"
wat
sh
[vagrant@localhost ~]$ echo "$0//a/"escaped quote"
-b"escaped quotesh
[vagrant@localhost ~]$ echo "$0//a/\escaped escape"
-bescaped escapesh
[vagrant@localhost ~]$ echo "$0//a/'escaped single quote"
-b'escaped single quotesh
[vagrant@localhost ~]$ echo "$0//a/''"
-b''sh
[vagrant@localhost ~]$ echo "$0//a/''''"
-b''''sh
[vagrant@localhost ~]$ echo "$0//a/'''"
> a'b}c"d
-b'''}"
a'bshcd
[vagrant@localhost ~]$ echo "$0//a/'''"
> w'x}y"z
-b'''}"
w'xshyz
[vagrant@localhost ~]$ echo "$0//a/''\"a test''"
> ^C
[vagrant@localhost ~]$ echo "$0//a/'''\"a test''"
-b'''"a test''sh
[vagrant@localhost ~]$ echo "$0//a/'''\"a test'$0'"
> ^C
[vagrant@localhost ~]$ echo "$0//a/\"a test'$0'"
> w}x"y
-b"a test'$0'}"
wshxy
[vagrant@localhost ~]$ echo "$0//a/\"a test'$0'"
-b"a test'$0'sh
[vagrant@localhost ~]$ echo "$0//a/\"a test'\'$0'"
> ^C
[vagrant@localhost ~]$ echo "$0//a/\"a test'\$0'"
-b"a test'-bash'sh
[vagrant@localhost ~]$