three questions on a Perl function - perl

Question

I am trying to use an existing Perl program, which includes the following function of GetItems. The way to call this function is listed in the following.
I have several questions for this program:
what does foreach my $ref (#_) aim to do? I think #_ should be related to the parameters passed, but not quite sure.
In my #items = sort { $a <=> $b } keys %items; the "items" on the left side should be different from the "items" on the right side? Why do they use the same name?
What does $items{$items[$i]} = $i + 1; aim to do? Looks like it just sets up the value for the hash $items sequentially.
$items = GetItems($classes, $pVectors, $nVectors, $uVectors);
######################################
sub GetItems
######################################
{
my $classes = shift;
my %items = ();
foreach my $ref (#_)
{
foreach my $id (keys %$ref)
{
foreach my $cui (keys %{$ref->{$id}}) { $items{$cui} = 1 }
}
}
my #items = sort { $a <=> $b } keys %items;
open(VAL, "> $classes.items");
for my $i (0 .. $#items)
{
print VAL "$items[$i]\n";
$items{$items[$i]} = $i + 1;
}
close VAL;
return \%items;
}

Answer 1

When you enter a function, #_ starts out as an array of (aliases to) all the parameters passed into the function; but the my $classes = shift removes the first element of #_ and stores it in the variable $classes, so the foreach my $ref (#_) iterates over all the remaining parameters, storing (aliases to) them one at a time in $ref.
Scalars, hashes, and arrays are all distinguished by the syntax, so they're allowed to have the same name. You can have a $foo, a #foo, and a %foo all at the same time, and they don't have to have any relationship to each other. (This, together with the fact that $foo[0] refers to #foo and $foo{'a'} refers to %foo, causes a lot of confusion for newcomers to the language; you're not alone.)
Exactly. It sets each element of %items to a distinct integer ranging from one to the number of elements, proceeding in numeric (!) order by key.

Answer 2

foreach my $ref (#_) loops through each hash reference passed as a parameter to GetItems. If the call looks like this:
$items = GetItems($classes, $pVectors, $nVectors, $uVectors);
then the loop processes the hash refs in $pVector, $nVectors, and $uVectors.
#items and %items are COMPLETELY DIFFERENT VARIABLES!! #items is an array variable and %items is a hash variable.
$items{$items[$i]} = $i + 1 does exactly as you say. It sets the value of the %items hash whose key is $items[$i] to $i+1.

Answer 3

Here is an (nearly) line by line description of what is happening in the subroutine
Define a sub named GetItems.
sub GetItems {
Store the first value in the default array #_, and remove it from the array.
my $classes = shift;
Create a new hash named %items.
my %items;
Loop over the remaining values given to the subroutine, setting $ref to the value on each iteration.
for my $ref (#_){
This code assumes that the previous line set $ref to a hash ref. It loops over the unsorted keys of the hash referenced by $ref, storing the key in $id.
for my $id (keys %$ref){
Using the key ($id) given by the previous line, loop over the keys of the hash ref at that position in $ref. While also setting the value of $cui.
for my $cui (keys %{$ref->{$id}}) {
Set the value of %item at position $cui, to 1.
$items{$cui} = 1;
End of the loops on the previous lines.
}
}
}
Store a sorted list of the keys of %items in #items according to numeric value.
my #items = sort { $a <=> $b } keys %items;
Open the file named by $classes with .items appended to it. This uses the old-style two arg form of open. It also ignores the return value of open, so it continues on to the next line even on error. It stores the file handle in the global *VAL{IO}.
open(VAL, "> $classes.items");
Loop over a list of indexes of #items.
for my $i (0 .. $#items){
Print the value at that index on it's own line to *VAL{IO}.
print VAL "$items[$i]\n";
Using that same value as an index into %items (which it is a key of) to the index plus one.
$items{$items[$i]} = $i + 1;
End of loop.
}
Close the file handle *VAL{IO}.
close VAL;
Return a reference to the hash %items.
return \%items;
End of subroutine.
}

Answer 4

I have several questions for this program:
What does foreach my $ref (#_) aim to do? I think #_ should be related to the parameters passed, but not quite sure.
Yes, you are correct. When you pass parameters into a subroutine, they automatically are placed in the #_ array. (Called a list in Perl). The foreach my $ref (#_) begins a loop. This loop will be repeated for each item in the #_ array, and each time, the value of $ref will be assigned the next item in the array. See Perldoc's Perlsyn (Perl Syntax) section about for loops and foreach loops. Also look at Perldoc's Perlvar (Perl Variables) section of General variables for information about special variables like #_.
Now, the line my $classes = shift; is removing the first item in the #_ list and putting it into the variable $classes. Thus, the foreach loop will be repeated three times. Each time, $ref will be first set to the value of $pVectors, $nVectors, and finally $uVectors.
By the way, these aren't really scalar values. In Perl, you can have what is called a reference. This is the memory location of the data structure you're referencing. For example, I have five students, and each student has a series of tests they've taken. I want to store all the values of each test in a hash keyed by the student's ID.
Normally, each entry in the hash can only contain a single item. However, what if this item refers to a list that contains the student's grades?
Here's the list of student #100's grade:
#grades = (100, 93, 89, 95, 74);
And here's how I set Student 100's entry in my hash:
$student{100} = \#grades;
Now, I can talk about the first grade of the year for Student #100 as $student{100}[0]. See the Perldoc's Mark's very short tutorial about references.
In my #items = sort { $a <=> $b } keys %items; the "items" on the left side should be different from the "items" on the right side? Why do they use the same name?
In Perl, you have three major types of variables: Lists (what some people call Arrays), Hashes (what some people call Keyed Arrays), and Scalars. In Perl, it is perfectly legal to have different variable types have the same name. Thus, you can have $var, %var, and #var in your program, and they'll be treated as completely separate variables1.
This is usually a bad thing to do and is highly discouraged. It gets worse when you think of the individual values: $var refers to the scalar while $var[3] refers to the list, and $var{3} refers to the hash. Yes, it can be very, very confusing.
In this particular case, he has a hash (a keyed array) called %item, and he's converting the keys in this hash into a list sorted by the keys. This syntax could be simplified from:
my #items = sort { $a <=> $b } keys %items;
to just:
my #items = sort keys %items;
See the Perldocs on the sort function and the keys function.
What does $items{$items[$i]} = $i + 1; aim to do? Looks like it just sets up the value for the hash $items sequentially.
Let's look at the entire loop:
foreach my $i (0 .. $#items)
{
print VAL "$items[$i]\n";
$items{$items[$i]} = $i + 1;
}
The subroutine is going to loop through this loop once for each item in the #items list. This is the sorted list of keys to the old %items hash. The $#items means the largest index in the item list. For example, if #items = ("foo", "bar", and "foobar"), then $#item would be 2 because the last item in this list is $item[2] which equals foobar.
This way, he's hitting the index of each entry in #items. (REMEMBER: This is different from %item!).
The next line is a bit tricky:
$items{$items[$i]} = $i + 1;
Remember that $item{} refers to the old %items hash! He's creating a new %items hash. This is being keyed by each item in the #items list. And, the value is being set to the index of that item plus 1. Let's assume that:
#items = ("foo", "bar", "foobar")
In the end, he's doing this:
$item{foo} = 1;
$item{bar} = 2;
$item{foobar} = 3;
1 Well, this isn't 100% true. Perl stores each variable in a kind of hash structure. In memory, $var, #var, and %var will be stored in the same hash entry in memory, but in positions related to each variable type. 99.9999% of the time, this matters not one bit. As far as you are concerned, these are three completely different variables.
However, there are a few rare occasions where a programmer will take advantage of this when they futz directly with memory in Perl.

Answer 5

I want to show you how I would write that subroutine.
Bur first, I want to show you some of the steps of how, and why, I changed the code.
Reduce the number of for loops:
First off this loop doesn't need to set the value of $items{$cui} to anything in particular. It also doesn't have to be a loop at all.
foreach my $cui (keys %{$ref->{$id}}) { $items{$cui} = 1 }
This does practically the same thing. The only real difference is it sets them all to undef instead.
#items{ keys %{$ref->{$id}} } = ();
If you really needed to set the values to 1. Note that (1)x#keys returns a list of 1's with the same number of elements in #keys.
my #keys = keys %{$ref->{$id}};
#items{ #keys } = (1) x #keys;
If you are going to have to loop over a very large number of elements then a for loop may be a good idea, but only if you have to set the value to something other than undef. Since we are only using the loop variable once, to do something simple; I would use this code:
$items{$_} = 1 for keys %{$ref->{$id}};
Swap keys with values:
On the line before that we see:
foreach my $id (keys %$ref){
In case you didn't notice $id was used only once, and that was for getting the associated value.
That means we can use values and get rid of the %{$ref->{$id}} syntax.
for my $hash (values %$ref){
#items{ keys %$hash } = ();
}
( $hash isn't a good name, but I don't know what it represents. )
3 arg open:
It isn't recommended to use the two argument form of open, or to blindly use the bareword style of filehandles.
open(VAL, "> $classes.items");
As an aside, did you know there is also a one argument form of open. I don't really recommend it though, it's mostly there for backward compatibility.
our $VAL = "> $classes.items";
open(VAL);
The recommend way to do it, is with 3 arguments.
open my $val, '>', "$classes.items";
There may be some rare edge cases where you need/want to use the two argument version though.
Put it all together:
sub GetItems {
# this will cause open and close to die on error (in this subroutine only)
use autodie;
my $classes = shift;
my %items;
for my $vector_hash (#_){
# use values so that we don't have to use $ref->{$id}
for my $hash (values %$ref){
# create the keys in %items
#items{keys %$hash} = ();
}
}
# This assumes that the keys of %items are numbers
my #items = sort { $a <=> $b } keys %items;
# using 3 arg open
open my $output, '>', "$classes.items";
my $index; # = 0;
for $item (#items){
print {$output} $item, "\n";
$items{$item} = ++$index; # 1...
}
close $output;
return \%items;
}
Another option for that last for loop.
for my $index ( 1..#items ){
my $item = $items[$index-1];
print {$output} $item, "\n";
$items{$item} = $index;
}
If your version of Perl is 5.12 or newer, you could write that last for loop like this:
while( my($index,$item) = each #items ){
print {$output} $item, "\n";
$items{$item} = $index + 1;
}