Perl function: accept hash, array, or scalar as parameter

Question

I have the following sub-routine.

 sub Function{
   my $ref = \($_[0]);
   if(ref($ref) eq 'SCALAR'){
   . . .
   }
}

I am trying to get it to work such that passing a list, scalar or hash to it, it converts that into a reference and depending upon whether that reference is of type ARRAY, SCALAR or HASH, different actions are performed.

It could be called as

Function(%a)
Function($a)
Function(@a)

How can I make $ref be a reference to whatever is passed to the Function? My current approach isn't working.

Dondi Michael Stroma · Accepted Answer · 2012-09-25 14:09:30Z

6

Function(%a)
Function($a)
Function(@a)

You can't do what you are asking. There is no way inside Function to determine whether an array or a hash was passed, because, in fact, you can't pass an array or a hash to a sub in the first place. You can only pass a list; the hashes and arrays are converted to lists in the sub calls above.

You should pass a reference to the function in the first place:

Function(\%a)
Function($a)
Function(\@a)

Then you can check what it is easily:

sub Function {
  my $param = shift;
  if (not ref $param) {
     ...
  } elsif (ref $param eq 'HASH') {
     ...
  } elsif (ref $param eq 'ARRAY') {
     ...
  }

answered Sep 25, 2012 at 14:09

Dondi Michael Stroma

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2 Comments

Anirudh Ramanathan Over a year ago

I understood that :) Thanks. But there is a way to pass the array or hash-ref, even if array or hash is passed by user. (stackoverflow.com/a/12583728/759019)

Dondi Michael Stroma Over a year ago

@DarkXphenomenon It's really quite hackish though. Why can't you pass a reference instead of a list? Are you working with legacy code? If so, I could see why you would want amon's solution, but, in new code, it is not a good design.

BenMorel · Accepted Answer · 2014-01-07 14:07:17Z

3

You can use a prototyped function. Although Perl 5 prototypes are a total mess, they can be used quite well in this case. As I understand them the prototype

sub Function (\[$@%]) {...}

should solve your problem, and the reference be in $_[0].

The big disadvantage is that the variable you pass has to start with the $, @, or % character, so you can't call Function with a constant argument directly:

 Function(1)

fails, but

 my $temp = 1;
 Function($temp)

works.

Your approach failes, because @_ is a list, and all elements are scalars.

edited Jan 7, 2014 at 14:07

BenMorel

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answered Sep 25, 2012 at 13:12

amon

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7 Comments

Anirudh Ramanathan Over a year ago

I understand why $_[0] failed previously. Thanks! I am using that Function() recursively and it is failing when I use the above prototype with the error Function called too early to check prototype when trying to recurse.

amon Over a year ago

@DarkXphenomenon then use subs 'Function' at the beginning of the module/file to force prototype checking

amon Over a year ago

@DarkXphenomenon Yes, but this scalar $_[0] is a reference to the original argument. You can view the contents with print Data::Dumper::Dump $_[0]

Anirudh Ramanathan Over a year ago

Dumper prints it right. The entire array. It is a ref to the original arg. But when I do ref($_[0]), it says SCALAR, even though it is an array I passed.

amon Over a year ago

@DarkXphenomenon For testing purposes, I used this one-liner: perl -MData::Dumper -e'sub Function (\[$@%]) {print $_[0], "\n"; print Dumper $_[0]} @x = (1, 2); Function(@x)' Output: ARRAY(0x8a5bcec) $VAR1 = [1, 2];. And ref is ARRAY.

|

user554546 · Accepted Answer · 2012-09-25 13:19:00Z

0

Why not just check to see if the argument passed in is a reference and act accordingly?

sub gimme_a_ref
{
  my $arg=shift;

  if(ref $arg ne ref "")
  {
    return $arg;
  }
  else
  {
    return \$arg;
  }
}

answered Sep 25, 2012 at 13:19

user554546

8 Comments

amon Over a year ago

given (ref $arg) {return $arg when [qw(ARRAY HASH CODE SCALAR)]; default {return \$arg}} is far cleaner.

pavel Over a year ago

or: sub gimme_a_ref { my $arg = shift; return ref $arg ? $arg : \$arg }

Anirudh Ramanathan Over a year ago

my $arg = shift; If I was passing a hash to the function, wouldn't the scalar $arg just contain the first key?

user554546 Over a year ago

@DarkXphenomenon - Well, what do you want to have happen if @_ > 1? Since => is treated the same as a comma, the subroutine wouldn't know if you intended to pass the elements of @_ as an array (that is not an array ref) or as a hash (that is not a hash ref) (with the exception of passing in an odd number of arguments, of course).

user554546 Over a year ago

@DarkXphenomenon - It sounds as though you might want to separate your functionality: one subroutine to take a list of arguments that will be treated as a hash, another to take a list of arguments that will be treated as an array, etc.

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