Hash with typed keys…
use v6;
class Foo {}
my Hash[Foo, Foo] $MAP;
my $f1 = Foo.new;
my $f2 = Foo.new;
$MAP{$f1} = $f2;
produces the error:
Invocant of method 'ASSIGN-KEY' must be an object instance of type 'Hash[Foo,Foo]', not a type object of type 'Hash[Foo,Foo]'. Did you forget a '.new'?
I find it misleading; what's the real error and what do I have to write instead?
I already tried the % sigil for the hash variable, that doesn't work, either.
In the way you have defined it, $MAP is actually a role. You need to instantiate (actually, pun) it:
class Foo {}
my Hash[Foo, Foo] $MAP;
my $map = $MAP.new;
my $f1 = Foo.new;
my $f2 = Foo.new;
$map{$f1} = $f2;
say $map;
Dead giveaway here was that classes can't be parametrized, roles do.
Also:
say $MAP.DEFINITE; # False
say $map.DEFINITE; # True
But actually the error message was pretty informative, up to and including the suggestion to use .new, as I do here.
We can shorten it down to:
class Foo {}
my %map = Hash[Foo, Foo].new ;
%map{Foo.new} = Foo.new;
%map.say;
By doing the punning from the definition, we don't need the $MAP intermediate class.
TL;DR JJ's answer is right, but the explanation left me confused. I currently view the problem you showed as an autovivification error/bug and/or LTA error message.
say my Any $Any; # (Any)
say my Hash $Hash; # (Hash)
say my Hash[Int] $Hash-Int; # (Hash[Int])
$Any<a> = 42; # OK
$Hash<a> = 42; # OK
$Hash-Int.new<a> = 42; # OK
$Hash-Int<a> = 42; # must be an object instance, not a type object
Imo this is a bug or pretty close to one.
A bug/problem applies for arrays too in the same scenario:
say my Any $Any; # (Any)
say my Array $Array; # (Array)
say my Array[Int] $Array-Int; # (Array[Int])
$Any[42] = 42; # OK
$Array[42] = 42; # OK
$Array-Int.new[42] = 42; # OK
$Array-Int[42] = 42; # Type check failed ... expected Array[Int] but got Array
If it's best considered notabug, then perhaps the error message should be changed. While I agree with JJ that the error message is actually on point (when you understand how raku works and figure out what's going on), I think it's nevertheless an LTA error message if we don't change raku(do) to dwim.
On the gripping hand, it's not obvious to me how one could best improve the error message. And now we have this SO. (cf my point about that in Is the ... error message LTA? in a recent answer I wrote.)
Another solution
I already tried the % sigil for the hash variable, that doesn't work, either.
JJ has provided a solution that initializes with a value with an explicit .new. But that drops the constraint from the variable. To retain it:
class Foo {}
constant FooFoo = Hash[Foo:D,Foo:D];
my %foo is FooFoo;
%foo{Foo.new} = Foo.new;
Ideally the constant wouldn't be needed, and perhaps one day it won't, but I think trait parsing is limited.
Related
Given the following code, it seems that I cannot iterate over a Buf if it had been assigned to a variable, unless I cast it to a list, even though it's not a lazy sequence. What gives?
my $file = open $path, bin => True;
$_.chr.say for $file.read: 8; # works well
my $test = $file.read: 8;
$_.chr.say for $test; # fails with "No such method 'chr' for invocant of type 'Buf[uint8]'"
$_.chr.say for $test.list; # works well
$test.is-lazy.say; # False
The reason it fails, is that:
my $test = $file.read: 8;
puts the Buf that is returned by $file.read into a Scalar variable, aka inside a container. And containerized is interpreted by for as itemized, to be considered a single item. So with:
.chr.say for $test;
you're calling the .chr method on the whole Buf, rather than on the individual elements.
There are a number of solutions to this:
make sure there's no container:
my $test := $file.read: 8;
This makes sure there is no container by binding the Buf.
make it look like an array
my #test := $file.read: 8;
Same as 1 basically, make the #test be an alias for the Buf. Note that this should also use binding, otherwise you'll get the same effect as you saw.
make it work like an Iterable
.chr.say for #$test;
By prefixing the # you're telling to iterate over it. This is basically syntactic sugar for the $test.list workaround you already found.
Re the $test.is-lazy.say, that is False for just about anything, e.g. 42.is-lazy.say; # False, so that doesn't tell you very much :-)
Suppose I have the following code:
my constant #suits = <Clubs Hearts Spades Diamonds>;
my constant #values = 2..14;
class Card {
has $.suit;
has $.value;
# order is mnemonic of "$value of $suit", i.e. "3 of Clubs"
multi method new($value, $suit) {
return self.bless(:$suit, :$value);
}
}
It defines some suits and some values and what it means to be a card.
Now, to build a deck, I essentially need to take the cross product of the suits and the values and apply that to the constructor.
The naiive approach to do this, would of course be to just iterate with a loop:
my #deck = gather for #values X #suits -> ($v, $c) {
take Card.new($v, $c);
}
But this is Raku, we have a cross function that can take a function as an optional argument!, so of course I'm gonna do that!
my #deck = cross(#values, #suits, :with(Card.new));
# Unexpected named argument 'with' passed
# in block <unit> at .\example.raku line 36
... wait no.
What about this?
my #deck = cross(#values, #suits):with(Card.new);
# Unexpected named argument 'with' passed
# in block <unit> at .\example.raku line 36
Still nothing. Reference maybe?
my #deck = cross(#values, #suits):with(&Card.new);
# ===SORRY!=== Error while compiling D:\Code\Raku/.\example.raku
# Illegally post-declared type:
# Card used at line 36
I read somewhere I can turn a function into an infix operator with []
my #deck = cross(#values, #suits):with([Card.new]);
# Unexpected named argument 'with' passed
# in block <unit> at .\example.raku line 36
That also doesn't work.
If classes are supposed to just be modules, shouldn't I then be able to pass a function reference?
Also why is it saying 'with' is that's unexpected? If I'm intuiting this right, what it's actually complaining about is the type of the input, rather than the named argument.
The error message is indeed confusing.
The :with parameter expects a Callable. Card.new is not a Callable. If you write it as :with( { Card.new($^number, $^suit) } ), it appears to work.
Note that I did not use $^value, $^suit, because they order differently alphabetically, so would produce the values in the wrong order. See The ^ twigil for more information on that syntax.
The error is LTA, this makes it a little bit better.
To get back to your question: you can find the code object that corresponds to Card.new with ^find_method. However, that will not work, as Card.new actually expects 3 arguments: the invocant (aka self), $value and $suit. Whereas the cross function will only pass the value and the suit.
The title of your question is “How do I take a reference to new?”, but that is not really what you want to do.
Raku being Raku, you can actually get a reference to new.
my $ref = Card.^lookup('new');
You can't use it like you want to though.
$ref(2,'Clubs'); # ERROR
The problem is that methods take a class or instance as the first argument.
$ref(Card, 2,'Clubs');
You could use .assuming to add it in.
$ref .= assuming(Card);
$ref(2,'Clubs');
But that isn't really any better than creating a block lambda
$ref = { Card.new( |#_ ) }
$ref(2,'Clubs');
All of these work:
cross( #values, #suits ) :with({Card.new(|#_)}) # adverb outside
cross( #values, #suits, :with({Card.new(|#_)}) ) # inside at end
cross( :with({Card.new(|#_)}), #values, #suits ) # inside at beginning
#values X[&( {Card.new(|#_)} )] #suits # cross meta-op with fake infix op
do {
sub new-card ($value,$suit) { Card.new(:$value,:$suit) }
#values X[&new-card] #suits
}
do {
sub with ($value,$suit) { Card.new(:$value,:$suit) }
cross(#values,#suits):&with
}
This code:
constant %what = { doesn't => 'change' };
%what = { will => "change" }
Should say something along the lines of "Cannot modify an immutable hash". However, it says:
Potential difficulties:
Useless use of hash composer on right side of hash assignment; did you mean := instead?
Positionals have pretty much the same problem, but the error is different. In this case it's about cannot modify an immutable, but an Str:
constant #what = <does not change>;
#what = <does change> # Cannot modify an immutable Str (does)
A Scalar works as expected. Is this a case of LTA error message, or is some container magic at work here that I'm missing?
This code:
constant %what = { doesn't => 'change' };
%what = { will => "change" }
Should say something along the lines of "Cannot modify an immutable hash".
Who says so? I mean this rhetorically, not rudely. I get why you think so but it's important to be careful with use of the word "should" because it implies some authority says so, eg the specification, or a design document, or someone's common sense, or whatever.
Per the current spec, and Rakudo implementation, what constant foo ... does is permanently (constantly) bind foo to some particular "value".
If that "value" is a container, then foo constantly refers to that container. (Yes, a container can be a "value", for some definition of "value" that is appropriate here.)
So your code above has changed the elements contained within that container, and that is, per spec, perfectly cromulent:
say %what; # {will => change}
In the meantime, the warning message legitimately mentions useless use of a hash constructor, plus it notes:
did you mean := instead?
If you try that:
constant %what = { doesn't => 'change' };
%what := { will => "change" }
You get:
Cannot use bind operator with this left-hand side
Because, as already established, %what is a compile time constant permanently bound to a hash created and initialized at compile time and that aspect -- the permanent binding of %what to that particular hash -- can't be changed during this program run.
Positionals have pretty much the same problem, but the error is different. In this case it's about cannot modify an immutable, but an Str:
constant #what = <does not change>;
#what = <does change> # Cannot modify an immutable Str (does)
That's a bit different. A constant declaration binds, regardless of whether you write = or :=. So the constant declarations are equivalent to:
constant %what := { doesn't => 'change' }
constant #what := <does not change>;
The first line binds %what to { doesn't => 'change' } which is a mutable Hash.
The second line binds #what to <does not change> which is an immutable List.
You could instead write:
constant #what = [<does not change>];
#what = <does change>;
say #what; # [does change]
A Scalar works as expected.
Not quite.
A scalar (lowercase, the generic term) does:
constant $scalar = 42;
$scalar = 99; # Cannot assign to an immutable value
Remembering that constant always binds, the above code parallels:
my $scalar := 42;
$scalar = 99; # Cannot assign to an immutable value
But a Scalar works the same as the other containers in this context:
constant $scalar = $ = 42;
$scalar = 99; # OK
(Don't write code like that unless you want to annoy people.)
Is this a case of LTA error message, or is some container magic at work?
That's a good question, and one I'm not going to try answer.
The Perl 6 Web site on functions says
Coercion types can help you to have a specific type inside a routine, but accept wider input. When the routine is called, the argument is automatically converted to the narrower type.
sub double(Int(Cool) $x) {
2 * $x
}
say double '21'; # 42
say double Any; # Type check failed in binding $x; expected 'Cool' but got 'Any'
Here the Int is the target type to which the argument will be coerced, and Cool is the type that the routine accepts as input.
But what is the point for the sub? Isn't $x just an Int? Why would you restrict the caller to implement Cool for the argument?
I'm doubly confused by the example because Int already is Cool. So I did an example where the types don't share a hierarchy:
class Foo { method foomethod { say 'foomethod' } }
class Bar {}
class Quux is Foo {
# class Quux { # compile error
method Bar { Bar.new }
}
sub foo(Bar(Foo) $c) {
say $c.WHAT; # (Bar)
# $c.foomethod # fails if uncommented: Method 'foomethod' not found for invocant of class 'Bar'
}
foo(Quux.new)
Here the invocant of foo is restricted to provide a Foo that can be converted to a Bar but foo cannot even call a method of Foo on $c because its type is Bar. So why would foo care that the to-be-coerced type is a Foo in the first place?
Could someone shed some light on this? Links to appropriate documentation and parts of the spec are appreciated as well. I couldn't find anything useful there.
Update Having reviewed this answer today I've concluded I had completely misunderstood what #musiKk was getting at. This was revealed most clearly in #darch's question and #musiKk's response:
#darch: Or is your question why one might prefer Int(Cool) over Int(Any)? If that's the case, that would be the question to ask.
#musiKk: That is exactly my question. :)
Reviewing the many other answers I see none have addressed it the way I now think it warrants addressing.
I might be wrong of course so what I've decided to do is leave the original question as is, in particular leaving the title as is, and leave this answer as it was, and instead write a new answer addressing #darch's reformulation.
Specify parameter type, with no coercion: Int $x
We could declare:
sub double (Int $x) { ... } # Accept only Int. (No coercion.)
Then this would work:
double(42);
But unfortunately typing 42 in response to this:
double(prompt('')); # `prompt` returns the string the user types
causes the double call to fail with Type check failed in binding $x; expected Int but got Str ("42") because 42, while looking like a number, is technically a string of type Str, and we've asked for no coercion.
Specify parameter type, with blanket coercion: Int() $x
We can introduce blanket coercion of Any value in the sub's signature:
sub double (Int(Any) $x) { ... } # Take Any value. Coerce to an Int.
Or:
sub double (Int() $x) { ... } # Same -- `Int()` coerces from Any.
Now, if you type 42 when prompted by the double(prompt('')); statement, the run-time type-check failure no longer applies and instead the run-time attempts to coerce the string to an Int. If the user types a well-formed number the code just works. If they type 123abc the coercion will fail at run-time with a nice error message:
Cannot convert string to number: trailing characters after number in '123⏏abc'
One problem with blanket coercion of Any value is that code like this:
class City { ... } # City has no Int coercion
my City $city;
double($city);
fails at run-time with the message: "Method 'Int' not found for invocant of class 'City'".
Specify parameter type, with coercion from Cool values: Int(Cool) $x
We can choose a point of balance between no coercion and blanket coercion of Any value.
The best class to coerce from is often the Cool class, because Cool values are guaranteed to either coerce nicely to other basic types or generate a nice error message:
# Accept argument of type Cool or a subclass and coerce to Int:
sub double (Int(Cool) $x) { ... }
With this definition, the following:
double(42);
double(prompt(''));
works as nicely as it can, and:
double($city);
fails with "Type check failed in binding $x; expected Cool but got City (City)" which is arguably a little better diagnostically for the programmer than "Method 'Int' not found for invocant of class 'City'".
why would foo care that the to-be-coerced type is a Foo in the first place?
Hopefully it's now obvious that the only reason it's worth limiting the coerce-from-type to Foo is because that's a type expected to successfully coerce to a Bar value (or, perhaps, fail with a friendly message).
Could someone shed some light on this? Links to appropriate documentation and parts of the spec are appreciated as well. I couldn't find anything useful there.
The document you originally quoted is pretty much all there is for enduser doc. Hopefully it makes sense now and you're all set. If not please comment and we'll go from there.
What this does is accept a value that is a subtype of Cool, and tries to transform it into an Int. At that point it is an Int no matter what it was before.
So
sub double ( Int(Cool) $n ) { $n * 2 }
can really be thought of as ( I think this is how it was actually implemented in Rakudo )
# Int is a subtype of Cool otherwise it would be Any or Mu
proto sub double ( Cool $n ) {*}
# this has the interior parts that you write
multi sub double ( Int $n ) { $n * 2 }
# this is what the compiler writes for you
multi sub double ( Cool $n ) {
# calls the other multi since it is now an Int
samewith Int($n);
}
So this accepts any of Int, Str, Rat, FatRat, Num, Array, Hash, etc. and tries to convert it into an Int before calling &infix:<*> with it, and 2.
say double ' 5 '; # 25
say double 2.5; # 4
say double [0,0,0]; # 6
say double { a => 0, b => 0 }; # 4
You might restrict it to a Cool instead of Any as all Cool values are essentially required to provide a coercion to Int.
( :( Int(Any) $ ) can be shortened to just :( Int() $ ) )
The reason you might do this is that you need it to be an Int inside the sub because you are calling other code that does different things with different types.
sub example ( Int(Cool) $n ) returns Int {
other-multi( $n ) * $n;
}
multi sub other-multi ( Int $ ) { 10 }
multi sub other-multi ( Any $ ) { 1 }
say example 5; # 50
say example 4.5; # 40
In this particular case you could have written it as one of these
sub example ( Cool $n ) returns Int {
other-multi( Int($n) ) * Int($n);
}
sub example ( Cool $n ) returns Int {
my $temp = Int($n);
other-multi( $temp ) * $temp;
}
sub example ( Cool $n is copy ) returns Int {
$n = Int($n);
other-multi( $n ) * $n;
}
None of them are as clear as the one that uses the signature to coerce it for you.
Normally for such a simple function you can use one of these and it will probably do what you want.
my &double = * * 2; # WhateverCode
my &double = * × 2; # ditto
my &double = { $_ * 2 }; # bare block
my &double = { $^n * 2 }; # block with positional placeholder
my &double = -> $n { $n * 2 }; # pointy block
my &double = sub ( $n ) { $n * 2 } # anon sub
my &double = anon sub double ( $n ) { $n * 2 } # anon sub with name
my &double = &infix:<*>.assuming(*,2); # curried
my &double = &infix:<*>.assuming(2);
sub double ( $n ) { $n * 2 } # same as :( Any $n )
Am I missing something? I'm not a Perl 6 expert, but it appears the syntax allows one to specify independently both what input types are permissible and how the input will be presented to the function.
Restricting the allowable input is useful because it means the code will result in an error, rather than a silent (useless) type conversion when the function is called with a nonsensical parameter.
I don't think an example where the two types are not in a hierarchical relationship makes sense.
Per comments on the original question, a better version of #musiKk's question "What is the point of coercions like Int(Cool)?" turned out to be:
Why might one prefer Int(Cool) over Int(Any)?
A corollary, which I'll also address in this answer, is:
Why might one prefer Int(Any) over Int(Cool)?
First, a list of various related options:
sub _Int_strong (Int $) {} # Argument must be Int
sub _Int_cool (Int(Cool) $) {} # Argument must be Cool; Int invoked
sub _Int_weak (Int(Any) $) {} # Argument must be Any; Int invoked
sub _Int_weak2 (Int() $) {} # same
sub _Any (Any $) {} # Argument must be Any
sub _Any2 ( $) {} # same
sub _Mu (Mu $) {} # Weakest typing - just memory safe (Mu)
_Int_strong val; # Fails to bind if val is not an Int
_Int_cool val; # Fails to bind if val is not Cool. Int invoked.
_Int_weak val; # Fails to bind if val is not Any. Int invoked.
_Any val; # Fails to bind if val is Mu
_Mu val; # Will always bind. If val is a native value, boxes it.
Why might one prefer Int(Cool) over Int(Any)?
Because Int(Cool) is slightly stronger typing. The argument must be of type Cool rather than the broader Any and:
Static analysis will reject binding code written to pass an argument that isn't Cool to a routine whose corresponding parameter has the type constraint Int(Cool). If static analysis shows there is no other routine candidate able to accept the call then the compiler will reject it at compile time. This is one of the meanings of "strong typing" explained in the last section of this answer.
If a value is Cool then it is guaranteed to have a well behaved .Int conversion method. So it will not yield a Method not found error at run-time and can be relied on to provide a good error message if it fails to produce a converted to integer value.
Why might one prefer Int(Any) over Int(Cool)?
Because Int(Any) is slightly weaker typing in that the argument can be of any regular type and P6 will just try and make it work:
.Int will be called on an argument that's passed to a routine whose corresponding parameter has the type constraint Int(...) no matter what the ... is. Provided the passed argument has an .Int method the call and subsequent conversion has a chance of succeeding.
If the .Int fails then the error message will be whatever the .Int method produces. If the argument is actually Cool then the .Int method will produce a good error message if it fails to convert to an Int. Otherwise the .Int method is presumably not a built in one and the result will be pot luck.
Why Foo(Bar) in the first place?
And what's all this about weak and strong typing?
An Int(...) constraint on a function parameter is going to result in either:
A failure to type check; or
An.Int conversion of the corresponding argument that forces it to its integer value -- or fails, leaving the corresponding parameter containing a Failure.
Using Wikipedia definitions as they were at the time of writing this answer (2019) this type checking and attempted conversion will be:
strong typing in the sense that a type constraint like Int(...) is "use of programming language types in order to both capture invariants of the code, and ensure its correctness, and definitely exclude certain classes of programming errors";
Currently weak typing in Rakudo in the sense that Rakudo does not check the ... in Int(...) at compile time even though in theory it could. That is, sub double (Int $x) {}; double Date; yields a compile time error (Calling double(Date) will never work) whereas sub double (Int(Cool) $x) {}; double Date; yields a run time error (Type check failed in binding).
type conversion;
weak typing in the sense that it's implicit type conversion in the sense that the compiler will handle the .Int coercion as part of carrying out the call;
explicit type conversion in the sense that the Int(...) constraint is explicitly directing the compiler to do the conversion as part of binding a call;
checked explicit type conversion -- P6 only does type safe conversions/coercions.
I believe the answer is as simple as you may not want to restrict the argument to Int even though you will be treating it as Int within the sub. say for some reason you want to be able to multiply an Array by a Hash, but fail if the args can't be treated as Int (i.e. is not Cool).
my #a = 1,2,3;
my %h = 'a' => 1, 'b' => 2;
say #a.Int; # 3 (List types coerced to the equivalent of .elems when treated as Int)
say %h.Int; # 2
sub m1(Int $x, Int $y) {return $x * $y}
say m1(3,2); # 6
say m1(#a,%h); # does not match
sub m2(Int(Cool) $x, Int(Cool) $y) {return $x * $y}
say m2('3',2); # 6
say m2(#a,%h); # 6
say m2('foo',2); # does not match
of course, you could also do this without the signature because the math operation will coerce the type automatically:
sub m3($x,$y) {return $x * $y}
say m3(#a,%h); # 6
however, this defers your type check to the inside of the sub, which kind of defeats the purpose of a signature and prevents you from making the sub a multi
All subtypes of Cool will be (as Cool requires them to) coerced to an Int. So if an operator or routine internal to your sub only works with Int arguments, you don't have to add an extra statement/expression converting to an Int nor does that operator/routine's code need to account for other subtypes of Cool. It enforces that the argument will be an Int inside of your sub wherever you use it.
Your example is backwards:
class Foo { method foomethod { say 'foomethod' } }
class Bar {}
class Quux is Bar {
method Foo { Foo.new }
}
sub foo(Foo(Bar) $c) {
#= converts $c of type Bar to type Foo
#= returns result of foomethod
say $c.WHAT; #-> (Foo)
$c.foomethod #-> foomethod
}
foo(Quux.new)
I'm studying dynamic/static scope with deep/shallow binding and running code manually to see how these different scopes/bindings actually work. I read the theory and googled some example exercises and the ones I found are very simple (like this one which was very helpful with dynamic scoping) But I'm having trouble understanding how static scope works.
Here I post an exercise I did to check if I got the right solution:
considering the following program written in pseudocode:
int u = 42;
int v = 69;
int w = 17;
proc add( z:int )
u := v + u + z
proc bar( fun:proc )
int u := w;
fun(v)
proc foo( x:int, w:int )
int v := x;
bar(add)
main
foo(u,13)
print(u)
end;
What is printed to screen
a) using static scope? answer=180
b) using dynamic scope and deep binding? answer=69 (sum for u = 126 but it's foo's local v, right?)
c) using dynamic scope and shallow binding? answer=69 (sum for u = 101 but it's foo's local v, right?)
PS: I'm trying to practice doing some exercises like this if you know where I can find these types of problems (preferable with solutions) please give the link, thanks!
Your answer for lexical (static) scope is correct. Your answers for dynamic scope are wrong, but if I'm reading your explanations right, it's because you got confused between u and v, rather than because of any real misunderstanding about how deep and shallow binding work. (I'm assuming that your u/v confusion was just accidental, and not due to a strange confusion about values vs. references in the call to foo.)
a) using static scope? answer=180
Correct.
b) using dynamic scope and deep binding? answer=69 (sum for u = 126 but it's foo's local v, right?)
Your parenthetical explanation is right, but your answer is wrong: u is indeed set to 126, and foo indeed localizes v, but since main prints u, not v, the answer is 126.
c) using dynamic scope and shallow binding? answer=69 (sum for u = 101 but it's foo's local v, right?)
The sum for u is actually 97 (42+13+42), but since bar localizes u, the answer is 42. (Your parenthetical explanation is wrong for this one — you seem to have used the global variable w, which is 17, in interpreting the statement int u := w in the definition of bar; but that statement actually refers to foo's local variable w, its second parameter, which is 13. But that doesn't actually affect the answer. Your answer is wrong for this one only because main prints u, not v.)
For lexical scope, it's pretty easy to check your answers by translating the pseudo-code into a language with lexical scope. Likewise dynamic scope with shallow binding. (In fact, if you use Perl, you can test both ways almost at once, since it supports both; just use my for lexical scope, then do a find-and-replace to change it to local for dynamic scope. But even if you use, say, JavaScript for lexical scope and Bash for dynamic scope, it should be quick to test both.)
Dynamic scope with deep binding is much trickier, since few widely-deployed languages support it. If you use Perl, you can implement it manually by using a hash (an associative array) that maps from variable-names to scalar-refs, and passing this hash from function to function. Everywhere that the pseudocode declares a local variable, you save the existing scalar-reference in a Perl lexical variable, then put the new mapping in the hash; and at the end of the function, you restore the original scalar-reference. To support the binding, you create a wrapper function that creates a copy of the hash, and passes that to its wrapped function. Here is a dynamically-scoped, deeply-binding implementation of your program in Perl, using that approach:
#!/usr/bin/perl -w
use warnings;
use strict;
# Create a new scalar, initialize it to the specified value,
# and return a reference to it:
sub new_scalar($)
{ return \(shift); }
# Bind the specified procedure to the specified environment:
sub bind_proc(\%$)
{
my $V = { %{+shift} };
my $f = shift;
return sub { $f->($V, #_); };
}
my $V = {};
$V->{u} = new_scalar 42; # int u := 42
$V->{v} = new_scalar 69; # int v := 69
$V->{w} = new_scalar 17; # int w := 17
sub add(\%$)
{
my $V = shift;
my $z = $V->{z}; # save existing z
$V->{z} = new_scalar shift; # create & initialize new z
${$V->{u}} = ${$V->{v}} + ${$V->{u}} + ${$V->{z}};
$V->{z} = $z; # restore old z
}
sub bar(\%$)
{
my $V = shift;
my $fun = shift;
my $u = $V->{u}; # save existing u
$V->{u} = new_scalar ${$V->{w}}; # create & initialize new u
$fun->(${$V->{v}});
$V->{u} = $u; # restore old u
}
sub foo(\%$$)
{
my $V = shift;
my $x = $V->{x}; # save existing x
$V->{x} = new_scalar shift; # create & initialize new x
my $w = $V->{w}; # save existing w
$V->{w} = new_scalar shift; # create & initialize new w
my $v = $V->{v}; # save existing v
$V->{v} = new_scalar ${$V->{x}}; # create & initialize new v
bar %$V, bind_proc %$V, \&add;
$V->{v} = $v; # restore old v
$V->{w} = $w; # restore old w
$V->{x} = $x; # restore old x
}
foo %$V, ${$V->{u}}, 13;
print "${$V->{u}}\n";
__END__
and indeed it prints 126. It's obviously messy and error-prone, but it also really helps you understand what's going on, so for educational purposes I think it's worth it!
Simple and deep binding are Lisp interpreter viewpoints of the pseudocode. Scoping is just pointer arithmetic. Dynamic scope and static scope are the same if there are no free variables.
Static scope relies on a pointer to memory. Empty environments hold no symbol to value associations; denoted by word "End." Each time the interpreter reads an assignment, it makes space for association between a symbol and value.
The environment pointer is updated to point to the last association constructed.
env = End
env = [u,42] -> End
env = [v,69] -> [u,42] -> End
env = [w,17] -> [v,69] -> [u,42] -> End
Let me record this environment memory location as AAA. In my Lisp interpreter, when meeting a procedure, we take the environment pointer and put it our pocket.
env = [add,[closure,(lambda(z)(setq u (+ v u z)),*AAA*]]->[w,17]->[v,69]->[u,42]->End.
That's pretty much all there is until the procedure add is called. Interestingly, if add is never called, you just cost yourself a pointer.
Suppose the program calls add(8). OK, let's roll. The environment AAA is made current. Environment is ->[w,17]->[v,69]->[u,42]->End.
Procedure parameters of add are added to the front of the environment. The environment becomes [z,8]->[w,17]->[v,69]->[u,42]->End.
Now the procedure body of add is executed. Free variable v will have value 69. Free variable u will have value 42. z will have the value 8.
u := v + u + z
u will be assigned the value of 69 + 42 + 8 becomeing 119.
The environment will reflect this: [z,8]->[w,17]->[v,69]->[u,119]->End.
Assume procedure add has completed its task. Now the environment gets restored to its previous value.
env = [add,[closure,(lambda(z)(setq u (+ v u z)),*AAA*]]->[w,17]->[v,69]->[u,119]->End.
Notice how the procedure add has had a side effect of changing the value of free variable u. Awesome!
Regarding dynamic scoping: it just ensures closure leaves out dynamic symbols, thereby avoiding being captured and becoming dynamic.
Then put assignment to dynamic at top of code. If dynamic is same as parameter name, it gets masked by parameter value passed in.
Suppose I had a dynamic variable called z. When I called add(8), z would have been set to 8 regardless of what I wanted. That's probably why dynamic variables have longer names.
Rumour has it that dynamic variables are useful for things like backtracking, using let Lisp constructs.
Static binding, also known as lexical scope, refers to the scoping mechanism found in most modern languages.
In "lexical scope", the final value for u is neither 180 or 119, which are wrong answers.
The correct answer is u=101.
Please see standard Perl code below to understand why.
use strict;
use warnings;
my $u = 42;
my $v = 69;
my $w = 17;
sub add {
my $z = shift;
$u = $v + $u + $z;
}
sub bar {
my $fun = shift;
$u = $w;
$fun->($v);
}
sub foo {
my ($x, $w) = #_;
$v = $x;
bar( \&add );
}
foo($u,13);
print "u: $u\n";
Regarding shallow binding versus deep binding, both mechanisms date from the former LISP era.
Both mechanisms are meant to achieve dynamic binding (versus lexical scope binding) and therefore they produce identical results !
The differences between shallow binding and deep binding do not reside in semantics, which are identical, but in the implementation of dynamic binding.
With deep binding, variable bindings are set within a stack as "varname => varvalue" pairs.
The value of a given variable is retrieved from traversing the stack from top to bottom until a binding for the given variable is found.
Updating the variable consists in finding the binding in the stack and updating the associated value.
On entering a subroutine, a new binding for each actual parameter is pushed onto the stack, potentially hiding an older binding which is therefore no longer accessible wrt the retrieving mechanism described above (that stops at the 1st retrieved binding).
On leaving the subroutine, bindings for these parameters are simply popped from the binding stack, thus re-enabling access to the former bindings.
Please see the the code below for a Perl implementation of deep-binding dynamic scope.
use strict;
use warnings;
use utf8;
##
# Dynamic-scope deep-binding implementation
my #stack = ();
sub bindv {
my ($varname, $varval);
unshift #stack, [ $varname => $varval ]
while ($varname, $varval) = splice #_, 0, 2;
return $varval;
}
sub unbindv {
my $n = shift || 1;
shift #stack while $n-- > 0;
}
sub getv {
my $varname = shift;
for (my $i=0; $i < #stack; $i++) {
return $stack[$i][1]
if $varname eq $stack[$i][0];
}
return undef;
}
sub setv {
my ($varname, $varval) = #_;
for (my $i=0; $i < #stack; $i++) {
return $stack[$i][1] = $varval
if $varname eq $stack[$i][0];
}
return bindv($varname, $varval);
}
##
# EXERCICE
bindv( u => 42,
v => 69,
w => 17,
);
sub add {
bindv(z => shift);
setv(u => getv('v')
+ getv('u')
+ getv('z')
);
unbindv();
}
sub bar {
bindv(fun => shift);
setv(u => getv('w'));
getv('fun')->(getv('v'));
unbindv();
}
sub foo {
bindv(x => shift,
w => shift,
);
setv(v => getv('x'));
bar( \&add );
unbindv(2);
}
foo( getv('u'), 13);
print "u: ", getv('u'), "\n";
The result is u=97
Nevertheless, this constant traversal of the binding stack is costly : 0(n) complexity !
Shallow binding brings a wonderful O(1) enhanced performance over the previous implementation !
Shallow binding is improving the former mechanism by assigning each variable its own "cell", storing the value of the variable within the cell.
The value of a given variable is simply retrieved from the variable's
cell (using a hash table on variable names, we achieve a
0(1) complexity for accessing variable's values!)
Updating the variable's value is simply storing the value into the
variable's cell.
Creating a new binding (entering subs) works by pushing the old value
of the variable (a previous binding) onto the stack, and storing the
new local value in the value cell.
Eliminating a binding (leaving subs) works by popping the old value
off the stack into the variable's value cell.
Please see the the code below for a trivial Perl implementation of shallow-binding dynamic scope.
use strict;
use warnings;
our $u = 42;
our $v = 69;
our $w = 17;
our $z;
our $fun;
our $x;
sub add {
local $z = shift;
$u = $v + $u + $z;
}
sub bar {
local $fun = shift;
$u = $w;
$fun->($v);
}
sub foo {
local $x = shift;
local $w = shift;
$v = $x;
bar( \&add );
}
foo($u,13);
print "u: $u\n";
As you shall see, the result is still u=97
As a conclusion, remember two things :
shallow binding produces the same results as deep binding, but runs faster, since there is never a need to search for a binding.
The problem is not shallow binding versus deep binding versus
static binding BUT lexical scope versus dynamic scope (implemented either with deep or shallow binding).