After reading the Raku documentation, I only found this for undefining a variable. I believe that in Raku there are differences between assignment and binding.
Defining and undefining a scalar is easy.
> my $n
(Any)
> $n.defined
False
> $n = 3
3
> $n.defined
True
> $n = Nil
(Any)
> $n.defined
False
When the variable is binded, it's not possible.
> my $k := 5
5
> $k := Nil
===SORRY!=== Error while compiling:
Cannot use bind operator with this left-hand side
at line 2
------> <BOL>⏏<EOL>
> $k = Nil
Cannot assign to an immutable value
in block <unit> at <unknown file> line 1
For arrays or hashes, I can empty it, but the variable is still defined.
For functions, when you define a function with sub, you cannot undefine it, but you can with an anonymous function.
> my &pera = -> $n { $n + 2}
-> $n { #`(Block|140640519305672) ... }
> &pera = Nil
(Callable)
> &pera.defined
False
> my &pera = -> $n { $n + 2}
-> $n { #`(Block|140640519305672) ... }
> &pera = Nil
(Callable)
> &pera.defined
False
> sub foo ($n) { $n + 1}
&foo
> &foo.defined
True
> &foo = Nil
Cannot modify an immutable Sub (&foo)
in block <unit> at <unknown file> line 1
So what's the difference between assignment and binding?
How can I undefine a variable?
Lots of different issues to discuss here.
> my $k := 5;
> $k := Nil;
Cannot use bind operator
This first problem is the Raku REPL. cf your last SO. Have you tried CommaIDE or repl.it?
Your code is perfectly valid:
my $k := 5;
$k := Nil;
say $k; # Nil
Moving on:
my $k := 5;
$k = Nil;
Cannot assign to an immutable value
This is different. After binding 5 to $k, the $k = Nil code is attempting to assign a value into a number. Only containers[1] support assignment. A number isn't a container, so you can't assign into it.
Clarifying some cases you mentioned but didn't explicitly cover:
my #foo;
#foo := Nil;
Type check failed in binding; expected Positional...
While scalar variables (ones with a $ sigil) will bind to any value or container, an # sigil'd variable will only bind to a Positional container such as an Array. (Likewise a % to an Associative such as a Hash.)
Not only that, but these containers are always defined. So they still return True for .defined even if they're empty:
my #foo := Array.new; # An empty array
say #foo.elems; # 0 -- zero elements
say #foo.defined; # True -- despite array being empty
Now assigning Nil to an array:
my #foo;
#foo = Nil;
say #foo; # [(Any)]
If a declaration of an # sigil'd variable doesn't bind it to some explicit Positional type it is instead bound to the default choice for an # variable. Which is an Array with a default element value of Any.
The #foo = Nil; statement above assigns a Nil value into the first element of #foo. The assignment of a value into a non-existing element of a multi-element container means a new Scalar container pops into existence and is bound to that missing element before assignment continues. And then, because we're assigning a Nil, and because a Nil denotes an absence of a value, the Array's default value ((Any)) is copied into the Scalar instead of the Nil.
On to the sub case...
sub foo {}
&foo = {} # Cannot modify an immutable Sub (&foo)
&foo := {} # Cannot use bind operator ...
While a sub foo declaration generates an &foo identifier, it is deliberately neither assignable nor bindable. If you want a Callable variable, you must declare one using ordinary variable declaration.
Unbind or undefine a variable
You can't unbind variables in the sense of leaving them bound to nothing at all. (In other words, Raku avoids the billion dollar mistake.) In some cases you can rebind variables.
In some cases you can bind or assign an undefined value to a variable. If it's not a scalar variable then that's like the # variable example covered above. The scalar cases are considered next.
An example of the binding case:
my $foo := Any;
say $foo.defined; # False
say $foo; # (Any)
say $foo.VAR.WHAT; # (Any)
We'll see what the .VAR is about in a moment.
The assignment case:
my $foo = Any;
say $foo.defined; # False
say $foo.WHAT; # (Any)
say $foo.VAR.WHAT; # (Scalar)
It's important to understand that in this case the $foo is bound to a Scalar, which is a container, which is most definitely "defined", for some definition of "defined", despite appearances to the contrary in the say $foo.defined; line.
In the say $foo.WHAT; line the Scalar remains hidden. Instead we see an (Any). But the (Any) is the type of the value held inside the Scalar container bound to $foo.
In the next line we've begun to pierce the veil by calling .VAR.WHAT on $foo. The .VAR gets the Scalar to reveal itself, rather than yielding the value it contains. And so we see the type Scalar.
But if you call .defined on that Scalar it still insists on hiding!:
my $foo;
say $foo.VAR.defined; # False
say $foo.VAR.DEFINITE; # True
(The only way to force Raku to tell you the ultimate truth about its view of definiteness is to call the ultimate arbiter of definiteness, .DEFINITE.)
So what are the rules?
The compiler will let you assign or bind a given new value or container to a variable, if doing so is valid according to the original variable declaration.
Assigning or binding an undefined value follows the same rules.
Binding
All variables must be bound by the end of their declaration.
If a variable's declaration allows an undefined value to be bound/assigned to that variable, then the variable can be undefined in that sense. But in all other circumstances and senses variables themselves can never be "unbound" or "undefined".
Binding is about making a variable correspond to some container or value:
Variables with # and % sigils must be bound to a container (default Array and Hash respectively).
Variables with the $ sigil must be bound to either a container (default Scalar) or a value.
Variables with the & sigil must be bound to a Callable value or a Scalar constrained to contain a Callable value. (The & sigil'd variable that's visible as a consequence of declaring a sub does not allow rebinding or assignment.)
Assignment
Assignment means copying a value into a container.
If a variable is bound to a container, then you can assign into it, provided the assignment is allowed by the compiler according to the original variable declaration.
If a variable is not bound to a container, then the compiler will reject an assignment to it.
Scalar variables
If you use a scalar container as if it were a value, then you get the value that's held inside the container.
Binding a value (defined or undefined) to a scalar variable will mean it will stop acting as a container. If you then try to assign to that variable it won't work. You'd need to rebind it back to a container.
Footnotes
[1] In this answer I've used the word "container" to refer to any value that can serve as a container for containing other values. For example, instances of Array, Hash, or Scalar.
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
}
multi sub infix:<*>( Numeric $i, Block $b ) { &$b($_) for ^($i.Int); }
3 * { .say };
Yields
Useless use of "*" in expression "3 * { .say }" in sink context
How do I get rid of that and make my operator work? I know I could assign it to $ or something else, but I don't want that.
Add this line to the start of your code:
proto sub infix:<*> ( | --> Nil ) {*}
See my answer to Impossible to put a map in sink context for a little on the --> Nil part of this (along with a boatload of irrelevant stuff too) including Larry's 2012 comment:
--> Nil seems like pretty good documentation of a procedure done only for its side-effects
I thought that in the top level of a program that will end and will leave would behave the same way, since there is only one big outer scope to exit/leave from. I thought that either one would be a good way to check a variable's final value.
But with will end it acts like the variable has never been initialized:
my $foo will end { put "Final value for \$foo is '$_'"} = 'bar';
put "\$foo is now '$foo'";
$foo ~= ' baz';
OUTPUT
$foo is now 'bar'
Use of uninitialized value $_ of type Any in string context.
Methods .^name, .perl, .gist, or .say can be used to stringify it to something meaningful.
in block at phaser_end.p6 line 1
Final value for $foo is ''
However, simply changing will end to will leave does what I would expect from either one:
my $foo will leave { put "Final value for \$foo is '$_'"} = 'bar';
put "\$foo is now '$foo'";
$foo ~= ' baz';
OUTPUT
$foo is now 'bar'
Final value for $foo is 'bar baz'
Why is there a difference in behavior here?
I am using Rakudo-Star 2017.07.
UPDATE
To get the effect that I'm expecting with will end, I have to use a separate END block:
END block:
my $foo = 'bar';
END { put "Final value for \$foo is '$foo'"};
put "\$foo is now '$foo'";
$foo ~= ' baz';
I guess the real question boils down to why does the END block behave differently than the will end block.
will end block:
my $foo will end { put "Final value for \$foo is '$_'"} = 'bar';
put "\$foo is now '$foo'";
$foo ~= ' baz';
Rewritten
Why does will end behave differently than will leave
It looks like will end has a bug similar to this old and now resolved bug for will begin.
Other than that, everything works as I would expect:
my $leave will leave { say ['leave:', $_, OUTERS::<$leave>] } = 'leave';
my $end will end { say ['end:', $_, OUTERS::<$end>] } = 'end';
my $begin will begin { say ['begin:', $_, OUTERS::<$begin>] } = 'begin';
END { say ['END:', $_, OUTERS::<$end>, $end] }
$_ = 999;
displays
[begin: (Any) (Any)]
[leave: leave leave]
[END: 999 end end]
[end: 999 end]
Is the scope for the block with will end different than the scope of the block with will leave?
They have the same outer lexical scope that corresponds to the UNIT scope.
They run in different dynamic scopes. Leave blocks run as part of leaving the enclosing block. End blocks run after the compiler has completely finished with your code and is cleaning up:
sub MAIN(#ARGS) {
...
# UNIT scope -- your code -- is about to be created and compiled:
$comp.command_line ...
# UNIT scope no longer exists. We compiled it, ran it, and left it.
...
# do all the necessary actions at the end, if any
if nqp::gethllsym('perl6', '&THE_END') -> $THE_END {
$THE_END()
}
}
why does the END block behave differently than the will end block?
Because you used $foo in one and $_ in the other.
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).