Is this some kind of regex within bash variable instantiation?

Question

What do those two assignations (i and C omitting the first one to void) do? Is it some kind of regex for the variable? I tried with bash, but so far there were no changes in the output of my strings after instantiating them with "${i//\\/\\\\}" or "\"${i//\"/\\\"}\""

C=''
for i in "$@"; do
    i="${i//\\/\\\\}"
    C="$C \"${i//\"/\\\"}\""
done

Oof. It looks like an attempt to store a list of DOS file paths in a single string by escaping backslashes and double quotes. A first step would be to simply use an array instead: `C=( "$@" )`. The backslashes should be left alone; if `C` knows about DOS file paths, it doesn't need them to be escaped, and if it doesn't, quoting them probably won't help. — chepner, Sep 07 '18 at 13:00

Mévatlavé Kraspek · Answer 1 · 2018-09-07T12:48:43.000

2

It's bash parameter expansions

it replace all backslashes by double backslashes :"${i//\\/\\\\}
it replace all \" by \\" : ${i//\"/\\\"}

Check http://wiki.bash-hackers.org/syntax/pe

edited Sep 07 '18 at 12:48

answered Sep 07 '18 at 12:43

Mévatlavé Kraspek

552
2
15

Tom Fenech · Accepted Answer · 2018-09-07T13:37:44.020

${i//\\/\\\\} is a slightly complicated-looking parameter expansion:

It expands the variable $i in the following way:

${i//find/replace} means replace all instances of "find" with "replace". In this case, the thing to find is \, which itself needs escaping with another \.
The replacement is two \, which each need escaping.

For example:

$ i='a\b\c'
$ echo "${i//\\/\\\\}"
a\\b\\c

The next line performs another parameter expansion:

find " (which needs to be escaped, since it is inside a double-quoted string)
replace with \" (both the double quote and the backslash need to be escaped).

It looks like the intention of the loop is to build a string C, attempting to safely quote/escape the arguments passed to the script. This type of approach is generally error-prone, and it would probably be better to work with the input array directly. For example, the arguments passed to the script can be safely passed to another command like:

cmd "$@" # does "the right thing" (quotes each argument correctly)

if you really need to escape the backslashes, you can do that too:

cmd "${@//\\/\\\\}" # replaces all \ with \\ in each argument

Is this some kind of regex within bash variable instantiation?

2 Answers2