[Mactoid]

I have just wasted way too much time at work trying to solve this question. I came across the problem while working on parsing script arguments. In a highly simplified form here is an example of the problem:

#!/bin/csh -f

set arg="-a"
if ( $arg =~ -*a* ) then # this works ok
echo "### -a =~ -a"
else
echo "### -a !~ -a"
endif

set arg="-f"
if ( $arg =~ -*f* ) then # looks the same, but doesn't work
echo "### -f =~ -f"
else
echo "### -f !~ -f"
endif

-f isn't the only value for arg that will cause an error, does anyone have any idea why the letter I use is causing me problems?

The "-" seems to be part of the problem, but I tried quoting the string and escaping the "-" and those didn't help.

Ideas anyone?

TIA

[ 06-21-2001: Message edited by: Mactoid ]

[howardm4]

possibly because the shell is interpreting the -f (which is a valid shell option) vs. the -a which isn't and thus passed on.

[Mactoid]

I think i'm understanding, the -f is actually being interpreted as an operator.
<3 minutes later>
Ah-ha. That was the problem. I fixed it by quoting $arg. Thanks for the input howardm4