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# 1  
Old 09-06-2012
Bash IFS
I am using bash and resetting IFS as below when reading the command line arguments. I do this so I can call my script as in Ex1.

Code:
Ex1: ./synt2d-ray3dmod.bash --xsrc=12/20/30

This allows me to split both sides so that when I do "shift"
I can get 12/20/30

What I do not understand is how calling Ex2 works also.

Ex2: ./synt2d-ray3dmod.bash --xsrc 12/20/30

Code:
OLDIFS="$IFS"
IFS="|="                # IFS controls splitting. Split on "|" and "="
set -- $*               # Set the positional parameters to the command line arguments.
IFS="$OLDIFS"

narg="$#"

while [ "$#" -gt 0 ]
do

  case "$1" in

  # name of output segy file
  "--ofl")
    shift
    val_ofl="${1}"
    hasArg_ofl="true"
  ;;

  # number ot time samples
  "--nt")
    shift
    val_nt="${1}"
    hasArg_nt="true"
  ;;

  # x-positions of sources
  "--xsrc")
    shift
    val_xSrc="${1}"
    hasArg_xSrc="true"
  ;;

  *)
    value_errLst="$arg_errLst ${1}"
    hasArg_errLst="true"
  ;;

  esac

  shift                 # Skip ahead to the next argument

done

# 2  
Old 09-06-2012
IFS works both ways. The shell splits strings apart on IFS, and when you put together a string with $*, it also puts them back together with IFS.

So, when you do ./myscript --xsrc 1234, $* actually turns it into --xsrc|1234. You just don't see it because you're letting the shell split it right back apart afterwards.

To prevent the shell from splitting something, you put it in quotes, so try this:

Code:
OLDIFS="$IFS"
IFS="|="                # IFS controls splitting. Split on "|" and "="
echo "$*"
set -- $*               # Set the positional parameters to the command line arg$
IFS="$OLDIFS"

Code:
$ ./script --arg=1234 # One argument

--arg=1234

$ ./script --arg 1234 # Two arguments

--arg|1234

$

So you see, the first time it comes in as 1 argument and nothing gets inserted by $*.

The second time it comes as two arguments, and gets stuck together into one string with | between.

IFS splits on both | and =, so set -- $* splits it apart where appropriate either way.
Last edited by Corona688; 09-06-2012 at 03:37 PM..
# 3  
Old 09-06-2012
This is not a "BASH thing", by the way. Any Bourne shell does this.
# 4  
Old 09-06-2012
There is another thing I am unsure about. It is when I am resetting IFS. My dilemma is what happens when I use the code I attached and the following

Code:
OLDIFS="$IFS"
IFS="|="                # IFS controls splitting. Split on "|" and "="
set -- $*               # Set the positional parameters to the command line arguments.

narg="$#"

while [ "$#" -gt 0 ]
do

  case "$1" in

  # name of output segy file
  "--ofl")
    shift
    val_ofl="${1}"
    hasArg_ofl="true"
  ;;

  # number ot time samples
  "--nt")
    shift
    val_nt="${1}"
    hasArg_nt="true"
  ;;

  # x-positions of sources
  "--xsrc")
    shift
    val_xSrc="${1}"
    hasArg_xSrc="true"
  ;;

  *)
    value_errLst="$arg_errLst ${1}"
    hasArg_errLst="true"
  ;;

  esac

  shift                 # Skip ahead to the next argument

done

IFS="$OLDIFS"

---------- Post updated at 02:06 PM ---------- Previous update was at 02:01 PM ----------

Originally my thinking was to reset IFS after the while (after processing all arguments). However, doing
Code:
OLDIFS="$IFS"
IFS="|="                # IFS controls splitting. Split on "|" and "="
set -- $*               # Set the positional parameters to the command line arg$
IFS="$OLDIFS"

while [ "$#" -gt 0 ]
do

...

does not seem to pose problems. What's happening?
# 5  
Old 09-06-2012
Changing IFS won't change splitting that's already happened. All the splitting for arguments happens in the third line, and gets saved into $1 $2 etc. The content and order of the $1 $2 ... variables won't change unless you do another set --.

I'd put the fourth line right below the first three just so you don't forget and try and split something later, and get weird results.
Last edited by Corona688; 09-06-2012 at 04:13 PM..
This User Gave Thanks to Corona688 For This Post:
kristinu
# 6  
Old 09-06-2012
Also, doing echo $* after set removes all = and replaces them with |

Code:
echo $*
set -- $*
echo $*

results is

Code:
./synt2d-ray3dmod.bash --xsrc=12/20/30 --verbose=3 --ofl=test.sgy

--xsrc=12/20/30|--verbose=3|--ofl=test.sgy
--xsrc|12/20/30|--verbose|3|--ofl|test.sgy

# 7  
Old 09-06-2012
That's not the code you ran. You must have been running echo "$*" quotes and all, otherwise you'd have never seen the pipes.

$* after the set didn't remove the equals, it was already gone. If you ran ./script a b --arg=c, $* on line 3 wopuld be the string a|b|--arg=c and split apart on "|=" into the arguments "a", "b", "--arg", and "c" The equals sign is already gone, deleted.

So when you do $* after that, it just smashes $1 $2 ... all together with | inbetween: a|b|--arg|c

Now remember, $* always, always does this, even if you don't quote it. Smilie You just don't see it without quotes, because the shell splits it. It'd see a|b|--arg|c, give "a" as echo's first argument, "b" as echo's second argument, "--arg" as echo's third argument, and "c" as echo's fourth argument. (echo doesn't split the arguments; that's the shell's job.) echo is not controlled by IFS, and will just put spaces inbetween...
Last edited by Corona688; 09-06-2012 at 04:52 PM..
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