How does this Perl grep work to determine the union of several hashes?

Question

I don't understand the last line of this function from Programming Perl 3e.

Here's how you might write a function that does a kind of set intersection by returning a list of keys occurring in all the hashes passed to it:

@common = inter( \%foo, \%bar, \%joe );
sub inter {
    my %seen;
    for my $href (@_) {
        while (my $k = each %$href) {
            $seen{$k}++;
        }
    }
    return grep { $seen{$_} == @_ } keys %seen;
}

I understand that %seen is a hash which maps each key to the number of times it was encountered in any of the hashes provided to the function.

Answer 1

grep will take a list passed to it (in this case, every element seen in any of the hashrefs); and return a list of only those elements where the expression in the block is true (locally setting $_ variable to each element in the list).

Let's look at how that expression is evaluated:

• @_ is an array of all the parameters passed to the subroutine - in our case a list of hash references passed in.

• In $seen{$_} == @_ expression that list is forced into a scalar context (due to ==).

• When used in a scalar context, a list evaluates to the number of elements in a list - in the example call above, to 3, since 3 hashrefs were passed in.

So, for each key in %seen (e.g. each key seen in any of N hashrefs); the expression $seen{$_} == @_ is numerically comparing the # of times the element was seen in the hashes to the total number of hashes - it's only going to be equal, of course, if the element is in ALL the hashes that were passed in, and thus a member of the intersection we want.

So, to sum up the analysis, the grep will return a list of all keys that occur in EVERY hash (aka occur N times where N is the # of hashes). E.g. an intersection.

Answer 2

grep block list 

This will apply block to each element of list in turn, the element is aliased as $_. If the block returns true, the element is added to the returned array.

in this case:

grep { $seen{$_} == @_ } keys %seen

The block is $seen{$_} == @_ , which compares the value of the seen hash against @_ . @_ is evaluated in scalar context and thus returns the number of elements in the @_ array. @_ represents the arguments to the current function. In this case ( \%foo, \%bar, \%joe ), which returns 3 in scalar context. Our list is keys %seen, which is an array containing all the keys present in %seen.

equivalent english statements:

• "give me a list of all the keys from %seen where the value associated with that key is equal to the number of elements passed to this function"
• "give me a list of all the keys from %seen where the value associated with that key is 3"
• "give me a list of all the keys from %seen that have value 3, ie all the keys from %seen that are present in each of the 3 hashrefs passed to this function"

Answer 3

The purpose of the function is to find the elements that appear in all the hashes passed to it.

The last line greps the list returned from keys %seen. To determine if a given key appears in all the hashes that were passed to the function, we can compare that key's value in %seen to the number of arguments to inter.

In the grep block, $_ is set to each element of the keys list, and tested for some condition.

An array in scalar context evaluates to its length. @_ is the array of arguments passed into the subroutine. And the == operator puts its operands in scalar context, so we can just compare the value of $seen{$_} to the length @_. If they're the same, then that key appeared in all the hashes.