Re: [bug-gawk] Passing Parameters to Gawk Executable Script

[Davide Brini]

On Sat, 1 Mar 2014 04:51:30 -0800 (PST), Mitchell Pryor
<address@hidden> wrote:

> Thanks for your response.  I'm running this script in Fedora 17 Linux, by
> the way. I understand that the operating system is involved in this
> process, but I'm not sure why you would say awk has nothing to do with
> it.  If ARGC says there are 3 arguments passed to the script does that
> not indicate that the operating system invoked awk and passed 3
> arguments?  Also, awk (gawk) is "seeing" the 'num' variable, but
> labelling it not 'legal'.  I would think an awk developer would have some
> idea why this would happen. But I guess I'm destined to continue my
> journey in the dark.  Thanks again.

That's the problem, ARGC is *NOT* 3. In fact there's no ARGC at all at the
time awk fails.

Basically the system is running your script as if it were invoked as:

awk '-v num=$1 -f' yourcode

(single quotes to pack everything into a single argument; try it and see,
"yourcode" can be anything). 

Awk bails out while still parsing its command line (as built by the system
out of the #! line).

It doesn't even get to the point where actual awk code would be run (and
thus concepts like ARGC etc. would start to make sense).

Look at the error message closely:

awk: fatal: ` num' is not a legal variable name

See, it tries to interpret " num" (with a leading space) as a variable
name. It's the command line parser that fails; the awk code in the script
does not matter.

-- 
D.