Re: Bash $@ Parameter Variable Breaking Out of Strings

[Reuti]

> Am 22.03.2016 um 15:47 schrieb Adam Danischewski 
> <adam.danischewski@gmail.com>:
> 
> I noticed an issue using the parameter built-in variable $@ breaking
> out of contained strings when utilized in functions. 
> 
> For example, consider the following bash script: m.bsh 
> #!/bin/bash
> echo "$#"
> while getopts d: OPTION "$@"; do
>  case "$OPTION" in
>    d)
>      echo "the optarg is ${OPTARG##*=}, optind is ${OPTIND}"
>      [[ -z "${OPTARG}" ]] && echo "Let's set the null byte as the delim."
>     ;;
>  esac
> done
> exit 0
> 
> $ alias t1='_() { var=$@; /tmp/m.bsh -d "clarify ${var}"; }; _'
> $ t1 hi there 
> 2
> the optarg is clarify hi there, optind is 3  
>  ### Correctly considers the text as a single string argument 
>      containing a space. 
> 
> $ alias t2='_() { /tmp/m.bsh -d "clarify $@"; }; _'

alias t2='_() { $HOME/m.bsh -d "clarify $*"; }; _'

might help.

For "$@" the man page says: If the double-quoted expansion occurs within a 
word, the expansion of  the first  parameter is joined with the beginning part 
of the original word [...]

-- Reuti


> $ t2 hi there 
> 3
> the optarg is clarify hi, optind is 3
>  ### Incorrectly breaks the argument array variable out as separate 
>      single string arguments. 
> 
> I noticed another interesting occurrence as well, I'm not sure if they are 
> related, to variable names: 
> 
> function update() { 
> local -i VAR=45
> VAR+=-1
> VAR+=$1
> echo $VAR
> } 
> 
> $ VAR2=2
> $ update VAR2 
>   47 
> 
> $ VAR=3
> $ update VAR 
>   88 ### !?  
> 
>