Good call on the letters versus characters. But as far as the prototype passing, I was certain that it enforces a pass by reference. Notice that I have to dereference it after the passage. So the hope is that there is no copying.
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sub sort_and_print_hash_keys (\%) {..
and how you eventually used the it like so:
sort_and_print_hash_keys(%letters); # you passed a hash variable not a
+ hash ref.
".. But as far as the prototype passing, I was certain that it enforces a pass by reference..."
If I may suggest, you will do well yielding the wisdom of jwkrahn, as regard the usage of prototype for this reason:
When you use a reference prototype, like "\$", "\@", "\%"
"...those symbols don't actually say that you must pass in a scalar reference, an array reference, and a hash reference. Rather, they say you must pass in a scalar variable, an array variable, and a hash variable. That means that the compiler insists upon seeing a properly notated variable of the given type, complete with "$", "@", or "%" in that slot. You must not use a backslash. The compiler silently supplies the backslash for you... "
from Far More than Everything You've Ever Wanted to Know about Prototypes in Perl -- by Tom Christiansen -- by liverpole, under subheading Problems with Reference Prototypes
Hope this helps.
If you tell me, I'll forget.
If you show me, I'll remember.
if you involve me, I'll understand.
--- Author unknown to me
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Notice that I have to dereference it after the passage. So the hope is that there is no copying.
In your code you have:
sub sort_and_print_hash_keys (\%) {
my %hash = %{shift()};
foreach (sort keys %hash) {
print "$hash{$_} ";
}
print "\n";
}
Which is copying the entire hash. If you didn't want to copy the hash you could do it like this:
sub sort_and_print_hash_keys (\%) {
my $hash = shift;
foreach (sort keys %$hash) {
print "$hash->{$_} ";
}
print "\n";
}
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