CODEWITHSUNDEEP
perlundefinedmatchingfunction-calls
PypeBros
PypeBros

Reputation: 2646

Perl function argument disappears. Why?

I'm building a stress-system in Perl for correcting students programming assignments. I built a function check(<boolean>,<congrats-message>,<blame-message>) that is supposed to help me along the way.

When I invoke it as 

check(get_option('content-type') eq "text/html","good type", "bad type");

everything is fine. But if I dare to do

check(get_option('content-type') =~ m:text/html:i, "good type", "bad type");

it breaks when the regexp doesn't find a match. Actually it is then equivalent to a check("good type", "bad type"). I use only $_[0], $_[1] etc. in check() function, and it's not like I had 'undef' as first argument: I can really catch the bug with die unless $#_ == 2.

What's happening ? I worked it around with 

check((get_option('content-type') =~ m:text/html:i && 1), "good type", "bad type");

but I'd love to understand the what and why of this odd situation.

--

sub check {
  if ($_[0]) {
    $okay++;
    print STDERR "^_^ $_[1] ($okay)\n";
  } else {
    print STDERR ">_< $_[2]\n";
  }
  return $_[0];
}

Upvotes: 2

Views: 122

Answers (1)

Quentin
Quentin

Reputation: 943108

eq always returns true or false, but =~ (when in list context) returns "a list consisting of the subexpressions matched by the parentheses in the pattern,".

When you put a list inside another list (e.g. the list of matches inside the list of arguments to the subroutine), they get merged.

If there are 0 matches, you get the 2 other arguments in the list.

If there are 2 matches, you get those 2 matches then the 2 other arguments for a total of 4 arguments.

Your code depends on there being exactly three arguments.

Upvotes: 6

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