4 ms·

I read both the Camel book (Programming Perl) and the slimmer one (Learning Perl) but I still found perl 5's split between by-value data structures (@foo/$foo[0

I read both the Camel book (Programming Perl) and the slimmer one (Learning Perl) but I still found perl 5's split between by-value data structures (@foo/$foo[0], %foo/$foo{"bar"}) and by-reference data structures ($foo->[0], $foo->{"bar"}) really confusing. Ruby and Python managed to avoid needing to differentiate these cases, they would just silently dereference things as needed, and from what I can tell Perl basically had this split for backward compatibility with perl 4 which didn't even support nested data structures. Steve Yegge has a whole section of his Perl rant talking about this (https://sites.google.com/site/steveyegge2/ancient-languages-perl https://sites.google.com/site/steveyegge2/ancient-languages-...). The sigils only make it more confusing because they really don't help you here. $foo is either a reference to @foo or a reference to an element of @foo depending on whether it is immediately followed by an index ("[0]") or if there is a dereference operator ("->") in between. And you only need the dereference operator if you're going to need to get a nested data structure. So like you can't just do $foo[2][4] (as I recall, it's been years) you have to do $foo->[2][4]. If you just need to go one level in you can do either $foo[2] or $foo->[2] depending on your whims. It's kind of madness.

codeflo 4y ago
First, I agree that using Perl today feels like using something from a different era. Having sigils at all feels so antiquated to me now. But the mechanics of the sigil thing never confused me. Either because it was explained to me very well, or possibly -- Larry Wall is a linguist, after all -- because there's an analogy to grammatical cases, which English largely doesn't have anymore, but my native language does. The sigil depends on what you want to get out of the expression, not what you put in. So @foo would name the whole array, while $foo[2] would name a single element from it. > If you just need to go one level in you can do either $foo[2] or $foo->[2] depending on your whims. It's kind of madness. You misremember, but I don't blame you, it's a very unorthodox rule. These are very different expressions. $foo[2] accesses the second (third depending on how you count) element of @foo. @foo is a "value array", whenever possible, I tended to use those because I found them more convenient and intuitive. For programmers coming from JavaScript or Python, and people building more complex data structures, using a reference array might be more intuitive. Those are stored in scalar variables, like $foo, but to disambiguate with the syntax for accessing @foo, you have to explicitly dereference it with $foo->[2].
- eduction 4y ago
  Ah ok I think I remember now: my @foo = (1,2); $foo[1]; #2 my $foo_r = \@foo $foo_r->[1]; #2 my @bar = (1,2,$foo_r); $bar[2]->[1]; #2 Re the sigils, you're right, I would just add, $ is for scalar and a scalar iirc can be /either/ a value or a reference, that's why $foo_r and $foo[1] above both start with $. That's what I found unhelpful, it would have been nice to have a distinct sigil for refs (but presumably Larry was out of available symbols at that point!).
  - TheDesolate0 4y ago
- messe 4y ago
  > there's an analogy to grammatical cases Best exemplified by Lingua Romana Perligata: https://metacpan.org/dist/Lingua-Romana-Perligata/view/lib/Lingua/Romana/Perligata.pm https://metacpan.org/dist/Lingua-Romana-Perligata/view/lib/L...

codeflo 4y ago

First, I agree that using Perl today feels like using something from a different era. Having sigils at all feels so antiquated to me now. But the mechanics of the sigil thing never confused me. Either because it was explained to me very well, or possibly -- Larry Wall is a linguist, after all -- because there's an analogy to grammatical cases, which English largely doesn't have anymore, but my native language does. The sigil depends on what you want to get out of the expression, not what you put in. So @foo would name the whole array, while $foo[2] would name a single element from it. > If you just need to go one level in you can do either $foo[2] or $foo->[2] depending on your whims. It's kind of madness. You misremember, but I don't blame you, it's a very unorthodox rule. These are very different expressions. $foo[2] accesses the second (third depending on how you count) element of @foo. @foo is a "value array", whenever possible, I tended to use those because I found them more convenient and intuitive. For programmers coming from JavaScript or Python, and people building more complex data structures, using a reference array might be more intuitive. Those are stored in scalar variables, like $foo, but to disambiguate with the syntax for accessing @foo, you have to explicitly dereference it with $foo->[2].

eduction 4y ago
Ah ok I think I remember now: my @foo = (1,2); $foo[1]; #2 my $foo_r = \@foo $foo_r->[1]; #2 my @bar = (1,2,$foo_r); $bar[2]->[1]; #2 Re the sigils, you're right, I would just add, $ is for scalar and a scalar iirc can be /either/ a value or a reference, that's why $foo_r and $foo[1] above both start with $. That's what I found unhelpful, it would have been nice to have a distinct sigil for refs (but presumably Larry was out of available symbols at that point!).
- TheDesolate0 4y ago
messe 4y ago
> there's an analogy to grammatical cases Best exemplified by Lingua Romana Perligata: https://metacpan.org/dist/Lingua-Romana-Perligata/view/lib/Lingua/Romana/Perligata.pm https://metacpan.org/dist/Lingua-Romana-Perligata/view/lib/L...