Re: [bug-gawk] What is wrong with getline from both stdin and a file from ARGV?

[Andrew J. Schorr]

On Wed, Nov 28, 2018 at 07:51:28PM -0600, Peng Yu wrote:
> > The problem with your second command, as I already explained, is that it 
> > opens
> > the file twice: the first time because it's supplied on the command-line as 
> > a
> > source of data, and the second time because you explicitly accessed it by
> > calling "getline < ARGV[1]".
> 
> I am not sure this explanation makes sense.

Trust me -- it makes sense. You have confused yourself because
stdin is irrelevant in your example. Please go back and reread my last
message slowly...

> Why the first getline in the following example read from /dev/fd/63?
> It is to read from stdin, shouldn't it be different from ARGV[1]?
> 
> $ printf '%s\n' 1 2 3 | awk -e 'BEGIN { getline; print getline <
> ARGV[1]; print  }' <(printf '%s\n' {a..c})
> 0
> a
> awk: (FILENAME=/dev/fd/63 FNR=1) warning: close of fd 63
> (`/dev/fd/63') failed (Bad file descriptor)

No -- it does not read from stdin, because you supplied a data file
on the command line. If there's a file on the command-line, then stdin
is ignored. For example:

bash-4.2$ echo hello | awk -F: 'NR == 1 {print FILENAME, $1}' /etc/passwd
/etc/passwd root

Do you understand now?

> Also, the following code works just fine. It would be strange to be
> able to read ARGV[2] but not ARGV[1]
> $ printf '%s\n' 1 2 3 | awk -e 'BEGIN { getline; print getline <
> ARGV[2]; print  }' <(printf '%s\n' {a..c})  <(printf '%s\n' {A..C})

Try printing out the data returned by your getline calls, and then maybe you
will understand.

Regards,
Andy