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c has asked for the wisdom of the Perl Monks concerning the following question:

The title pretty much describes what I am seeing. Actually, the return from $? is 0. I read over this which had made me think that $? would hold more info than the $! that I had been using, but to no avail.

I'm doing a pretty simple rcs checkout, move a file over another and then checking back in:

my $checkout = "co -q -l";
my $checkin = "ci -u -q -m\"Modification of $domain\"";

## check out file from rcs
system("$checkout $revdns")
  or warn "Problems with checking out $revdns : $!\n";

## move the tmp file into its place
move("$tmp_revdns","$revdns")
  or warn "Problems moving $tmp_revdns to $revdns : $!\n";

## check the new file into rcs
system("$checkin $revdns")
  or warn "Problems with checking in $revdns : $?\n";

print "inverse dns records created!\n";

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I get the warn statement with the first system call. Any ideas on why I would receive this?

humbly -c

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•Re: Illegal Seek from system() with no output from $?
by merlyn (Sage) on Mar 27, 2002 at 05:12 UTC
    The value of $! is not useful or defined when it's not checked immediately after a failed system call (which doesn't include system, which is really a macro for a combination of fork/exec/wait steps). So, you're getting bogus results from a bogus operation -- not surprising. {grin}

    -- Randal L. Schwartz, Perl hacker

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Re: Illegal Seek from system() with no output from $?
by vek (Prior) on Mar 27, 2002 at 07:34 UTC
    system returns 0 on success and 1 on failure. Whenever you use system you should be using and to check for failure instead of or: 
    system("$checkout $revdns")
   and warn "Problems with checking out $revdns : $!\n";

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Re: Illegal Seek from system() with no output from $?
by lestrrat (Deacon) on Mar 27, 2002 at 02:56 UTC

    I think system() returns the exit status of the command ( i.e. $? ) -- so if your command executed properly, then it would most likely return a 0, which is a logical "false" for perl. So you need to warn when: 

       system( "$your_command" );
   if( $? != 0 ) {
      warn "something's wrong";
   }

   ## or
   ##    ( system( "$cmd" ) == 0 ) or warn "the sky is falling!";
   ##

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    I don't know about the illegal seek stuff, but I used to get a lot of that when I would have bogus errors ( i.e. my bug ) in the code. It's probably just that $! was never meant to be set when you checked for it ( however, I'm no expert on this subject, so please take that statement with a grain of salt )

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