Re: Re: Re: Re: Test for standard input

[mikfire]

Sorry, I cannot resist. Your code is ummmm doing lots of things that I do not think are what you intend.

$ARGS = @ARGV;
if ( -t and $ARGS eq 0 ) {
}

The more idiomatic way of saying this is

die "No input found\n" if ( -t and @ARGV == 0 );

I would also like to point out that you are testing the wrong kind of equality in your code -- you used the string equality test ( eq ) instead of the numeric test ( == ). In this case it worked, but it will someday drive you nuts trying to hunt this kind of bug down.

$names++, shift if $ARGV[0] eq "-l";

If I have figured this out correctly, you are trying to parse some command line options. I would strongly recommend using Getopt::Long, a core module that does this job very well.

I am assuming you really mean to shift the argument you just parsed out of @ARGV. Unfortunately, shift with no arguments assumes you mean @_, which is not @ARGV. Again, it likely works in this case, but I really doubt this is your intended affect.

The same holds true for the next line.

@ARGV = "-" unless @ARGV;

This code will never be executed. I am assuming it is a hold-over from your previous problems. The 'if' at the top of the program makes certain that @ARGV has to have something in it.

The assignment back to @ARGV is a little.... ugly IMHO. It took me several tries to figure out what it is doing. I would likely write this expression as

exit 0 unless ( grep { -T or $_ eq "-" } @ARGV );

but this is likely more a stylistic thing.

Oh, good use of the -w flag. It appears, though, you forgot to use strict. It will seem a pain at first, but it will really help as your programs get longer and more complex.

Mik
mikfire

[BUU]

<pedantry>shift in a non subroutine context shifts from @ARGV.</pedantry>

[mikfire]

Damn. I have been programming perl for years now and I did not know this. Thanks!

Mik
mikfire