Re: [Help-bash] Fastest way to join an array into a string

[Stephane Chazelas]

2019-08-26 13:47:36 -0500, Peng Yu:
[...]
>             echo -n "$1"

You can't use echo for arbitrary data. Depending on the
environment and how bash was built, that will either
- output that -n
- fail if $1 is -n, -E, -e or any combination like -neEneE
- fail if $1 contains backslashes

Use printf again: printf %s "$1"

>             shift

You'll get an error if $@ is the empty list (no argument).

>             if (($#)); then
>                 declare x

Why "declare x"

>                 printf "${separator//%/%%}%s" "$@"
[...]

You also need to escape \. You're also missing the newline
delimiter.

Note that if you're printing it, that means that if you need to
store it in a variable, you'll need to use command substitution,
which involves running an extra process and writing/reading the
data through a pipe.

Even if you use printf -v, like in:

join_into() {
  local -n _var="$1"
  local _sep="$2"
  shift 2
  if (($# > 1)); then
    sep=${sep//%/%}
    sep=${sep//\\/\\\\}
    printf -v _var "${separator//%/%%}%s" "${@:2}"
    _var=$1$_var
  else
    _var=$1
  fi
}

I find it's still orders of magnitude slower than the approach that uses
${*/#/$sep}.

On my system, it is significantly slower than invoking perl or python

like

result=$(perl -le 'print join shift, @ARGV' -- sep "$@")

or

result=$(python -c 'import sys; print(sys.argv[1].join(sys.argv[2:]))' sep "$@")

-- 
Stephane