Bash script in-braces weird escaping and pound symbol

Question

I'm reading a bash script function which takes an input variable with escape codes as first parameter and cycles on its characters, anyway I can't understand these lines:

local input="${1//\"/\\\"}" output="" i char within_code=0
    for ((i=0; i < ${#input}; ++i)); do
        char="${input:i:1}"                     # get current character

I guess my bash-foo isn't strong enough.

• What is "${1//\\"/\\\\\\"}" ? The first parameter should be $1 I think.
• What does the # in ${#input} do?

The rest I can, more or less, understand.

Answer 1

The expansions are explained in the bash manual Shell Parameter Expansion .

1. It substitutes all occurrences of " with \\" in $1 . Yes, $1 expands to the first positional parameter.
2. The ${#input} expands to the length of $input .