Dynamic/Static scope with Deep/Shallow binding (exercises)

Question

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!

Answer 1

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!

Answer 2

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).

Answer 3

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.