Re: Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Greg Wooledge
On Tue, Mar 22, 2016 at 11:35:46AM -0400, Adam Danischewski wrote: > function update() { > local -i VAR=45 > local -i TOPSECRET_NUMBER=50 ## No one can know about. > VAR+=${1} > echo $VAR > } > > $ update 0 > 45 > $ update TOPSECRET_NUMBER > 95 > > ## Successfully deduce that TOPSECRET_NUMBER is

Re: Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Greg Wooledge
Today's lesson is: intent. Always tell us what you are trying to do. In a script, that means writing comments. In an email, you can write plain text anywhere you want. It's helpful! Try it! On Tue, Mar 22, 2016 at 10:47:28AM -0400, Adam Danischewski wrote: > $ alias t2='_() { /tmp/m.bsh -d "cl

Re: Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Adam Danischewski
Thanks for the quick reply, I hit send by accident, the first part is clear in the documentation. On the second part, I still see a potential problem because Bash is allowing for access to the internal local function variables. E.g. function update() { local -i VAR=45 local -i TOPSECRET_NUMBER=50

Re: Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Chet Ramey
On 3/22/16 10:47 AM, Adam Danischewski wrote: > 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 > cas

Re: Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Adam Danischewski
I hit send by accident, disregard the first part - I had the info page open and didn't realize I had sent the email already, the "$*" is what I should be using. The second part is still of a mystery, again a quick typo where it states VAR2=2, it should read VAR2=3 to obtain 47. On Tue, Mar 22, 20

Re: Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Reuti
> Am 22.03.2016 um 15:47 schrieb Adam Danischewski > : > > 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

Bash $@ Parameter Variable Breaking Out of Strings

2016-03-22 Thread Adam Danischewski
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##*=}, o