6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
6.001 Notes,Section 12.3
Slide 12.3.1
Because this is the central part of the environment model,let's
look in very painful detail at an example of an evaluation,In
particular,let's look at the evaluation of (square 4) with
respect to the global environment,Here is the structure we start
with,Assume that x has already been bound to the value 4 by
some define expression in the environment,and that we
have created the definition of square as we just saw,it is
pointing to the indicated procedure object,
Now,we want to see is how the rules for the environment
model tell us how to get the value associated with squaring 4 in
this environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.2
This is a compound expression,so first we have to get the
values of the subexpressions with respect to the same
environment,4 is easy to evaluate,it's just 4,We also need to
get the value of square with respect to the global
environment,
Slide 12.3.3
..,and that just comes from applying the name rule,We look
up the binding for square in this environment,and it simply
points to that double bubble as shown,
Slide 12.3.4
Aha! We are applying a procedure,one of those double
bubbles,to a set of arguments,so our four-step rule now comes
into play,Step one,create a new frame,lets call it A,
Slide 12.3.5
Step two,convert that frame into an environment,by having its
enclosing environment pointer point to the same environment as
the environment pointer of the procedure object,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.6
In fact,we can create a nice notation to keep track of this,I can
link these two pointers together,to indicate that the enclosing
environment of Frame A comes from the application of the
indicated procedure object,
Slide 12.3.7
Step three,take the formal parameters of the procedure object
being applied,in this case x,and bind them within that new
frame to the corresponding procedure argument values,in this
case 4,
Slide 12.3.8
Now,with respect to that new environment,E1,evaluate the
body of the procedure being applied,So notice,evaluating
(square 4) with respect to one environment,has
reduced to evaluating a simpler expression,(* x x),with
respect to a new environment,
Slide 12.3.9
Now the same rules apply as before,This is a compound
expression,so we need to get the values of the subexpressions
with respect to E1,We start by getting the value of * with
respect to E1,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.10
Well,.,* certainly doesn't have a binding in the first frame of
E1,That frame came from the application of the procedure
object,and only formal parameters of the procedure are bound
there,So our rule says to go up the enclosing environment
pointer to the global environment and look for a binding in that
environment,
We didn't tell you this,but in fact the global environment
creates a bindings for all the built-in procedures,Thus,* is
bound to the primitive multiplication procedure in that
environment,Thus,our name rule looks up the value associated
with * and returns a pointer to this primitive procedure,
Slide 12.3.11
So in this case we do get a value associated with *,
Slide 12.3.12
Remember where we were,We were getting the values of the
subexpressions of (* x x) with respect to E1,We have the
value of the first subexpression,so what about the value of x
with respect to E1? This is just the name rule,so starting in E1,
look for a binding of this variable,Of course,we find such a
binding in the first frame of E1,so we get back the value
associated with x in that frame,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.13
In particular,we get back the value 4,Not the value 10 to
which x is bound in the global environment! Remember we
start in the first frame of E1 looking for a binding for this
variable,Since there is such a binding,it shadows the other
binding in the global environment,
Slide 12.3.14
And now we complete this process,We get the value of the
second x in the same way,Thus we are left with the application
of a primitive procedure to numbers,This just returns the value
16,which is the value of the last expression in the body of the
procedure being applied,Thus,this is the value returned for the
entire expression,
Although this was a long example,step back to notice how the
mechanistic rules of the environment model simply tell us how
to trace the evaluation of an expression with respect to an
environment,reducing it to simpler expressions,
Slide 12.3.15
Now,let's be slightly more daring! Having seen the application
of a simple procedure like square,let's look at something
that involves a little more work,In particular,let's assume that
we have defined square and inc-square,as shown
in this environment structure,
Slide 12.3.16
So lets evaluate (inc-square 4) with respect to the
global environment,and here is the environment structure
corresponding to that environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.17
As in the previous case,this is a compound expression,so we
need to first evaluate the subexpressions with respect to the
same environment,The value of inc-square,by the
name rule,is just the double bubble pointed to by that variable,
Slide 12.3.18
And as we saw before,our four-step rule kicks in,Step one:
create a frame,
Slide 12.3.19
Step two,turn it into an environment,by having the enclosing
environment pointer of the frame point to the environment
specified by the procedure that is being applied..,
Slide 12.3.20
..,and that we know is specified by the second part of the
double bubble of the procedure being used,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.21
Step three,take the formal parameter of this procedure and
create a binding for it in the new frame,to the value of the
argument passed in,
Slide 12.3.22
Step four,take the body of that procedure object and evaluate
it with respect to this new environment,that is,(+ 1
(square y)) with respect to E1,
Slide 12.3.23
Again,notice how we have reduced the evaluation of one
compound expression with respect to one environment to the
evaluation of a simpler compound expression with respect to
another environment,
As before,to evaluate this compound expression,we first need
the values of the subexpressions,The value of + with respect to
E1 is determined by the name rule,chasing up the environment
chain from E1 to the global environment to get the primitive
addition operation,The value of 1 is just a self-evaluation rule,
Slide 12.3.24
So all we have left to do is get the value of (square y)
with respect to E1,This is again a compound expression,being
evaluated with respect to this new environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.25
To complete this evaluation,we need to get the value of
(square y) with respect to E1,and here is a clean
version of the environment structure developed so far,
Slide 12.3.26
Well,let's do it in steps,We first get the values of each of the
subexpressions,now with respect to E1,First,what is the value
of square with respect to E1? This is a little different than
last time,We start in E1,Since there is no binding for
square there,we go up the environment chain to the global
environment,where we find the binding,and thus return,.,
Slide 12.3.27
..,the appropriate procedure,
Slide 12.3.28
The second subexpression is y and we need its value with
respect to E1,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.29
..,and our name rule says to simply look it up with respect to
this environment chain starting in E1,This,of course,returns
the value 4,
Slide 12.3.30
So we are set for our big rule again,We have the application of
a procedure,one of those double bubbles,to a set of values,
Step one says,drop a frame,
Slide 12.3.31
Extend that frame into an environment by having its enclosing
environment pointer match that specified by the procedure
object..,
Slide 12.3.32
..,which says we want this one,Notice an interesting point
here,This new environment E2 is scoped by the global
environment,not by E1,You might have thought this should
be E1,because that was where we were doing the evaluation,
but remember that rule says the enclosing environment is
specified by the procedure being applied,not by the frame in
which we are doing the evaluation,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.33
Step three,create a binding,that is,take the formal parameter
of the procedure,x,and bind it in that frame to the value
passed in,4,
Slide 12.3.34
And step four,with respect to that new environment,evaluate
the body of this procedure,(* x x),So we have reduced
this evaluation to multiplying x by itself within environment
E2,
Slide 12.3.35
So now we are almost done,The values of each of these
subexpression with respect to E2 are obtained by applying the
name rule,starting in the first frame of E2,We then apply the
primitive multiplication procedure to the value of 4 and 4,
Slide 12.3.36
That of course just returns 16,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.37
And remember what started all of this,That value then gets
passed back out to where we were when we started this
diagram,which is adding 1 to the value returned by this
procedure application,finally resulting in 17,
There are a lot of details here,but the goal was to show how the
rules of the environment model very mechanistically specify
exactly the order in which to evaluate expressions,and how to
look up the bindings of variables in the appropriate
environment in order to make sure expression has a legal value,
Slide 12.3.38
So here are the key points to take away from this example,The
main point is to see how the rules for the environment model
specify almost everything we need to know in order to
understand how expressions are evaluated,It doesn't quite do
all of it,in particular,it doesn't specify the pending operations,
as we saw in the last example,
The other two points to notice are summarized on the slide,
6.001 Notes,Section 12.4
Slide 12.4.1
Now that we are reasonably comfortable with the environment
model,let's go back and tackle the problem that started us off
on this discussion,Remember this example from the beginning
of lecture?
Counter,or something created by make-counter,was
an object that should count up from a number,Every time we
applied that procedure of no arguments,we would get as output
the next number in sequence,We want to understand how
evaluating this same expression could give rise to different
values at different times,And how evaluating the same
expression could give us different objects,with independent
behavior,
Let's see if the environment model helps explain this computation,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.2
Although there will be a lot of details in this example,you
should try to follow the high level view of how the model
explains the computation,
First,lets evaluate this definition of ca to be the value of
applying make-counter in the global environment,We
have shown our structure for the global environment,with a
binding for make-counter to a procedure object,as
would be obtained when we evaluate the definition on the
previous slide,
Slide 12.4.3
To do this,we know we need to get the value of applying
make-counter to 0,Make-counter is a
procedure so the next set of stages is something we have
already seen,We will drop a frame,extend it into an
environment by scoping it by the same environment as the
environment pointer of the procedure,and within that frame,
bind the formal parameters of the procedure to the arguments
passed in,Thus we get the structure shown,which is exactly
like our previous procedure applications,
Slide 12.4.4
Now watch carefully,We now have an evaluation of the body
of that procedure with respect to an environment,and what is
that body? It is,itself,a lambda,with no parameters and a
body that does a mutation and returns a value,
Slide 12.4.5
So we just apply our rule for lambdas,We create a double
bubble for the procedure object created by the lambda,The
code part of the object is just the formal parameter (in this case
nothing) and the body (in this case (set! n (+ n
1)) and then n),The key issue is where does the second
part of the double bubble go? What is the environment pointer
we want here?
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.6
Right! It points to E1,because this is the environment in which
we were evaluating the lambda expression,Notice that this
is a different structure than we have seen before,For the first
time,we have a procedure object whose environment pointer
points to a frame or environment other than the global
environment,And in a second we are going to see why that is
crucial,
Slide 12.4.7
Now remember,that red (that double bubble) is the value
actually returned by the evaluation of the body of the procedure
we applied,So that is the value returned by (make
counter 0),Therefore,we can complete our definition,
The binding for ca up in the global environment,since that is
where we were evaluating the define,is now created,and
that variable points to the object returned by applying make-
counter,which is that procedure object,
This is a useful structure,The variable ca,which is available
to us in the global environment,points to a procedure,much as
earlier things did,but this procedure has nested within it an internal environment,Its environment pointer points to
an environment that is scoped relative to the global environment,
Slide 12.4.8
So having created this procedure object associated with ca,
let's look at what happens when we apply it,when we evaluate
it with no arguments,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.9
Here's a recap of the environment structure,We have the global
environment with a binding for ca,which points to a
procedure object whose environment pointer points to a new
frame,or environment,E1,This frame is scoped by the global
environment,but contains within it its own local variable n,
Slide 12.4.10
The value of ca is just that procedure object,so we are going
to apply a procedure,and we know what rules to use,Drop a
frame,
Slide 12.4.11
Within that frame,bind the formal parameters of this procedure,
There aren't any,so there is nothing to put in the frame,
Slide 12.4.12
And relative to this new environment,evaluate the body of the
procedure,which says we are going to evaluate (set! n
(+ n 1)) and then n,with respect to E2,Notice the
structure here,We now have a frame,E2,that points to a frame
E1,that points to the global environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.13
First,we know what a set! does,but let's look at this
carefully,First,we will get the value of (+ n 1) with
respect to E2,This means we look up the value of n with
respect to E2,Since it is not bound there,we go up the
enclosing environment pointer,to find the binding in E1,This
gives us the value 0,Thus we add 0 to 1,to get 1,
Notice how that local frame has capture n for us,Given that,
we can now evaluate the set! with respect to E2,This says,
starting in E2,trace up the environment chain until we find a
binding for n,which we find in E1,We then change that
binding to the newly computed value,which is 1,
Slide 12.4.14
Having evaluated that part of the body of the procedure,we
then evaluate that next part,n,with respect to E2,Using the
same rules,we chase up the chain of environment pointers until
we find a binding for n,returning the value 1 as the value for
this expression,Since this is the last expression in the body of
the procedure,this is also the value we return for the application
of the procedure itself,
The key thing to note here is how that local frame,E1,captures
some state information that is accessible only by this procedure,
Calling ca gives us the ability to get this value of n,change or
mutate it,and then return that value,
Slide 12.4.15
Now let's see what happens if we call ca again,We should
expect to see the behavior of the value incrementing,so let's
evaluate (ca) again with respect to the global environment,
Shown is a replication of the environment structure left from
the last evaluation,with n having been mutated from 0 to 1 as
part of that process,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.16
The value of ca is still the same procedure,and since we are
applying a procedure,we use our four step rule,We drop a
frame,
Slide 12.4.17
We then scope it with the same environment pointer as the
procedure object being applied,thus E1 as we did before,
Slide 12.4.18
There are no parameters to bind in this frame,
Slide 12.4.19
So evaluating (ca) with respect to the global environment
reduces to evaluating the body,(set! n (+ n 1))
with respect to this new frame E3,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.20
We know what should happen in this case,We just did this!
The evaluation will again mutate the value of n to be one more
than its current value,thus changing the value from 1 to 2 in
this frame,
Slide 12.4.21
And having evaluated that part of the body,we now take the
second part,n,and evaluate that with respect to E3,Chasing up
the frame pointers finds the binding in E1,and returns that
value,2,Since this is the value of the last expression in the
body of ca,this is the value returned by the whole thing,
Thus we see how this local piece of state allows us to have a
procedure,which when evaluated in successive turns,returns a
different value,
Slide 12.4.22
So our environment model helps us understand how one
counter,ca,can have some local state which it can keep
mutating,Thus,it can return a different value each time it is
applied,
What happens when we call make-counter again,
starting from 0,but giving it a different name,
Slide 12.4.23
And just to recap,here is the environment structure we have to
this point,There is a variable ca pointing to the procedure
shown,with local state of n equal to 2,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.24
Since we want to apply make-counter to 0,we first
look up the value of make-counter with respect to the
global environment,Since it is a procedure,which is being to
applied to 0 our four-step rule applies,Step one,drop a frame,
Slide 12.4.25
Step two,convert it into an environment by scoping that frame
with the same environment pointer as the procedure being
applied,Note that this means in this case pointing to the global
environment,
Slide 12.4.26
Step three,bind the formal parameter of this procedure,n,to
the value passed in,0,Note that this is a different n than the
one we had before,This n lives in environment E4,The n we
had for ca lives in environment E1,so we have different
bindings for the same name in different environments,
Slide 12.4.27
Relative to this environment,evaluate the body of the
procedure,Remember the procedure we are applying,make-
counter,so we are going to evaluate that lambda
expression with respect to E4,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.28
And we know what evaluating a lambda should do,It
creates a procedure object,a double bubble,the code of which
has no parameter and a body that is the same as the earlier body
we created for ca,The key point is,where does the procedure
object's environment pointer point to? To the environment in
which the lambda is being evaluated,thus to E4,Note that
this procedure thus has a different scoping than the one created
for ca,It is going to capture E4,with that version of n,which
is different than the one we had before,
Slide 12.4.29
And finally,that procedure object is the value returned by
applying (make-counter 0),and therefore the
definition creates a binding for cb up in the global
environment to that new procedure object,
Look carefully at the structure we have here,We have ca
pointing to a procedure object that inherits a frame E1,We have
cb pointing to a similar procedure object that inherits a frame
E4,They have different bindings for the parameter n,and that
is going to allow these things to behave differently,
Slide 12.4.30
So lets finish our understanding of this process by looking at
what happens when we evaluate (cb)> within the global
environment,
Slide 12.4.31
Here,again,is the environment structure we have so far,We
have a global environment with a binding for ca to one
procedure object with its own frame,and a binding for cb to
another procedure object with its own frame,What happens
when we apply cb?
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.32
Well,the rules make it very clear,We take the procedure object
associated with the binding for cb,and we apply it,This will
drop a frame,scope it by using the same environment pointer as
the procedure object,which says E5 will be scoped by E4,Not
E1,not global environment,but E4,
Slide 12.4.33
This says that using exactly the same reasoning as before,the
value of n that will get mutated is the one that is seen from E5,
namely the one sitting in E4,So it gets changed by 1,and then
we return that value,of 1,
Slide 12.4.34
This says the value of the overall expression is just 1,
Slide 12.4.35
So what was the point of this long,drawn-out,exercise? Part of
it was to let you see how the rules for the environment model
evolve,They specify what happens during a computation,they
specify how expressions get meanings assigned to them through
a very mechanistic set of rules,
But the second point was to let us understand things that
involve mutation,especially mutation associated with
procedures,So think about what happened with our little
counter example,Here is the environment model we had,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.36
And notice that there is a structure associated with it,What does
ca look like? It's a name,bound in the global environment,
that points to a procedure that has as its environment pointer,a
pointer into a frame,in this case E1,that has some local state in
it,
The only way we can get at that value of n is through a
procedure whose environment pointer,through some chain of
frames,points into that frame,In this case,that is only through
ca,
So this structure or pattern is a common pattern,
Slide 12.4.37
And in fact when we evaluated make-counter a second
time and gave the result returned by it the name cb,we created
a similar pattern,but now with its own local state,Thus,each of
these names specifies a procedure that has associated with it
some information that belongs only to that procedure,
This kind of framework,that is,procedures that capture local
state and can manipulate local state,is going to be a very useful
programming tool,as we are going to see in the next few
lectures,
Slide 12.4.38
Here is a summary of the key points to take away from this
exercise,
6.001 Notes,Section 12.3
Slide 12.3.1
Because this is the central part of the environment model,let's
look in very painful detail at an example of an evaluation,In
particular,let's look at the evaluation of (square 4) with
respect to the global environment,Here is the structure we start
with,Assume that x has already been bound to the value 4 by
some define expression in the environment,and that we
have created the definition of square as we just saw,it is
pointing to the indicated procedure object,
Now,we want to see is how the rules for the environment
model tell us how to get the value associated with squaring 4 in
this environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.2
This is a compound expression,so first we have to get the
values of the subexpressions with respect to the same
environment,4 is easy to evaluate,it's just 4,We also need to
get the value of square with respect to the global
environment,
Slide 12.3.3
..,and that just comes from applying the name rule,We look
up the binding for square in this environment,and it simply
points to that double bubble as shown,
Slide 12.3.4
Aha! We are applying a procedure,one of those double
bubbles,to a set of arguments,so our four-step rule now comes
into play,Step one,create a new frame,lets call it A,
Slide 12.3.5
Step two,convert that frame into an environment,by having its
enclosing environment pointer point to the same environment as
the environment pointer of the procedure object,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.6
In fact,we can create a nice notation to keep track of this,I can
link these two pointers together,to indicate that the enclosing
environment of Frame A comes from the application of the
indicated procedure object,
Slide 12.3.7
Step three,take the formal parameters of the procedure object
being applied,in this case x,and bind them within that new
frame to the corresponding procedure argument values,in this
case 4,
Slide 12.3.8
Now,with respect to that new environment,E1,evaluate the
body of the procedure being applied,So notice,evaluating
(square 4) with respect to one environment,has
reduced to evaluating a simpler expression,(* x x),with
respect to a new environment,
Slide 12.3.9
Now the same rules apply as before,This is a compound
expression,so we need to get the values of the subexpressions
with respect to E1,We start by getting the value of * with
respect to E1,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.10
Well,.,* certainly doesn't have a binding in the first frame of
E1,That frame came from the application of the procedure
object,and only formal parameters of the procedure are bound
there,So our rule says to go up the enclosing environment
pointer to the global environment and look for a binding in that
environment,
We didn't tell you this,but in fact the global environment
creates a bindings for all the built-in procedures,Thus,* is
bound to the primitive multiplication procedure in that
environment,Thus,our name rule looks up the value associated
with * and returns a pointer to this primitive procedure,
Slide 12.3.11
So in this case we do get a value associated with *,
Slide 12.3.12
Remember where we were,We were getting the values of the
subexpressions of (* x x) with respect to E1,We have the
value of the first subexpression,so what about the value of x
with respect to E1? This is just the name rule,so starting in E1,
look for a binding of this variable,Of course,we find such a
binding in the first frame of E1,so we get back the value
associated with x in that frame,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.13
In particular,we get back the value 4,Not the value 10 to
which x is bound in the global environment! Remember we
start in the first frame of E1 looking for a binding for this
variable,Since there is such a binding,it shadows the other
binding in the global environment,
Slide 12.3.14
And now we complete this process,We get the value of the
second x in the same way,Thus we are left with the application
of a primitive procedure to numbers,This just returns the value
16,which is the value of the last expression in the body of the
procedure being applied,Thus,this is the value returned for the
entire expression,
Although this was a long example,step back to notice how the
mechanistic rules of the environment model simply tell us how
to trace the evaluation of an expression with respect to an
environment,reducing it to simpler expressions,
Slide 12.3.15
Now,let's be slightly more daring! Having seen the application
of a simple procedure like square,let's look at something
that involves a little more work,In particular,let's assume that
we have defined square and inc-square,as shown
in this environment structure,
Slide 12.3.16
So lets evaluate (inc-square 4) with respect to the
global environment,and here is the environment structure
corresponding to that environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.17
As in the previous case,this is a compound expression,so we
need to first evaluate the subexpressions with respect to the
same environment,The value of inc-square,by the
name rule,is just the double bubble pointed to by that variable,
Slide 12.3.18
And as we saw before,our four-step rule kicks in,Step one:
create a frame,
Slide 12.3.19
Step two,turn it into an environment,by having the enclosing
environment pointer of the frame point to the environment
specified by the procedure that is being applied..,
Slide 12.3.20
..,and that we know is specified by the second part of the
double bubble of the procedure being used,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.21
Step three,take the formal parameter of this procedure and
create a binding for it in the new frame,to the value of the
argument passed in,
Slide 12.3.22
Step four,take the body of that procedure object and evaluate
it with respect to this new environment,that is,(+ 1
(square y)) with respect to E1,
Slide 12.3.23
Again,notice how we have reduced the evaluation of one
compound expression with respect to one environment to the
evaluation of a simpler compound expression with respect to
another environment,
As before,to evaluate this compound expression,we first need
the values of the subexpressions,The value of + with respect to
E1 is determined by the name rule,chasing up the environment
chain from E1 to the global environment to get the primitive
addition operation,The value of 1 is just a self-evaluation rule,
Slide 12.3.24
So all we have left to do is get the value of (square y)
with respect to E1,This is again a compound expression,being
evaluated with respect to this new environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.25
To complete this evaluation,we need to get the value of
(square y) with respect to E1,and here is a clean
version of the environment structure developed so far,
Slide 12.3.26
Well,let's do it in steps,We first get the values of each of the
subexpressions,now with respect to E1,First,what is the value
of square with respect to E1? This is a little different than
last time,We start in E1,Since there is no binding for
square there,we go up the environment chain to the global
environment,where we find the binding,and thus return,.,
Slide 12.3.27
..,the appropriate procedure,
Slide 12.3.28
The second subexpression is y and we need its value with
respect to E1,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.29
..,and our name rule says to simply look it up with respect to
this environment chain starting in E1,This,of course,returns
the value 4,
Slide 12.3.30
So we are set for our big rule again,We have the application of
a procedure,one of those double bubbles,to a set of values,
Step one says,drop a frame,
Slide 12.3.31
Extend that frame into an environment by having its enclosing
environment pointer match that specified by the procedure
object..,
Slide 12.3.32
..,which says we want this one,Notice an interesting point
here,This new environment E2 is scoped by the global
environment,not by E1,You might have thought this should
be E1,because that was where we were doing the evaluation,
but remember that rule says the enclosing environment is
specified by the procedure being applied,not by the frame in
which we are doing the evaluation,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.33
Step three,create a binding,that is,take the formal parameter
of the procedure,x,and bind it in that frame to the value
passed in,4,
Slide 12.3.34
And step four,with respect to that new environment,evaluate
the body of this procedure,(* x x),So we have reduced
this evaluation to multiplying x by itself within environment
E2,
Slide 12.3.35
So now we are almost done,The values of each of these
subexpression with respect to E2 are obtained by applying the
name rule,starting in the first frame of E2,We then apply the
primitive multiplication procedure to the value of 4 and 4,
Slide 12.3.36
That of course just returns 16,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.3.37
And remember what started all of this,That value then gets
passed back out to where we were when we started this
diagram,which is adding 1 to the value returned by this
procedure application,finally resulting in 17,
There are a lot of details here,but the goal was to show how the
rules of the environment model very mechanistically specify
exactly the order in which to evaluate expressions,and how to
look up the bindings of variables in the appropriate
environment in order to make sure expression has a legal value,
Slide 12.3.38
So here are the key points to take away from this example,The
main point is to see how the rules for the environment model
specify almost everything we need to know in order to
understand how expressions are evaluated,It doesn't quite do
all of it,in particular,it doesn't specify the pending operations,
as we saw in the last example,
The other two points to notice are summarized on the slide,
6.001 Notes,Section 12.4
Slide 12.4.1
Now that we are reasonably comfortable with the environment
model,let's go back and tackle the problem that started us off
on this discussion,Remember this example from the beginning
of lecture?
Counter,or something created by make-counter,was
an object that should count up from a number,Every time we
applied that procedure of no arguments,we would get as output
the next number in sequence,We want to understand how
evaluating this same expression could give rise to different
values at different times,And how evaluating the same
expression could give us different objects,with independent
behavior,
Let's see if the environment model helps explain this computation,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.2
Although there will be a lot of details in this example,you
should try to follow the high level view of how the model
explains the computation,
First,lets evaluate this definition of ca to be the value of
applying make-counter in the global environment,We
have shown our structure for the global environment,with a
binding for make-counter to a procedure object,as
would be obtained when we evaluate the definition on the
previous slide,
Slide 12.4.3
To do this,we know we need to get the value of applying
make-counter to 0,Make-counter is a
procedure so the next set of stages is something we have
already seen,We will drop a frame,extend it into an
environment by scoping it by the same environment as the
environment pointer of the procedure,and within that frame,
bind the formal parameters of the procedure to the arguments
passed in,Thus we get the structure shown,which is exactly
like our previous procedure applications,
Slide 12.4.4
Now watch carefully,We now have an evaluation of the body
of that procedure with respect to an environment,and what is
that body? It is,itself,a lambda,with no parameters and a
body that does a mutation and returns a value,
Slide 12.4.5
So we just apply our rule for lambdas,We create a double
bubble for the procedure object created by the lambda,The
code part of the object is just the formal parameter (in this case
nothing) and the body (in this case (set! n (+ n
1)) and then n),The key issue is where does the second
part of the double bubble go? What is the environment pointer
we want here?
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.6
Right! It points to E1,because this is the environment in which
we were evaluating the lambda expression,Notice that this
is a different structure than we have seen before,For the first
time,we have a procedure object whose environment pointer
points to a frame or environment other than the global
environment,And in a second we are going to see why that is
crucial,
Slide 12.4.7
Now remember,that red (that double bubble) is the value
actually returned by the evaluation of the body of the procedure
we applied,So that is the value returned by (make
counter 0),Therefore,we can complete our definition,
The binding for ca up in the global environment,since that is
where we were evaluating the define,is now created,and
that variable points to the object returned by applying make-
counter,which is that procedure object,
This is a useful structure,The variable ca,which is available
to us in the global environment,points to a procedure,much as
earlier things did,but this procedure has nested within it an internal environment,Its environment pointer points to
an environment that is scoped relative to the global environment,
Slide 12.4.8
So having created this procedure object associated with ca,
let's look at what happens when we apply it,when we evaluate
it with no arguments,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.9
Here's a recap of the environment structure,We have the global
environment with a binding for ca,which points to a
procedure object whose environment pointer points to a new
frame,or environment,E1,This frame is scoped by the global
environment,but contains within it its own local variable n,
Slide 12.4.10
The value of ca is just that procedure object,so we are going
to apply a procedure,and we know what rules to use,Drop a
frame,
Slide 12.4.11
Within that frame,bind the formal parameters of this procedure,
There aren't any,so there is nothing to put in the frame,
Slide 12.4.12
And relative to this new environment,evaluate the body of the
procedure,which says we are going to evaluate (set! n
(+ n 1)) and then n,with respect to E2,Notice the
structure here,We now have a frame,E2,that points to a frame
E1,that points to the global environment,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.13
First,we know what a set! does,but let's look at this
carefully,First,we will get the value of (+ n 1) with
respect to E2,This means we look up the value of n with
respect to E2,Since it is not bound there,we go up the
enclosing environment pointer,to find the binding in E1,This
gives us the value 0,Thus we add 0 to 1,to get 1,
Notice how that local frame has capture n for us,Given that,
we can now evaluate the set! with respect to E2,This says,
starting in E2,trace up the environment chain until we find a
binding for n,which we find in E1,We then change that
binding to the newly computed value,which is 1,
Slide 12.4.14
Having evaluated that part of the body of the procedure,we
then evaluate that next part,n,with respect to E2,Using the
same rules,we chase up the chain of environment pointers until
we find a binding for n,returning the value 1 as the value for
this expression,Since this is the last expression in the body of
the procedure,this is also the value we return for the application
of the procedure itself,
The key thing to note here is how that local frame,E1,captures
some state information that is accessible only by this procedure,
Calling ca gives us the ability to get this value of n,change or
mutate it,and then return that value,
Slide 12.4.15
Now let's see what happens if we call ca again,We should
expect to see the behavior of the value incrementing,so let's
evaluate (ca) again with respect to the global environment,
Shown is a replication of the environment structure left from
the last evaluation,with n having been mutated from 0 to 1 as
part of that process,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.16
The value of ca is still the same procedure,and since we are
applying a procedure,we use our four step rule,We drop a
frame,
Slide 12.4.17
We then scope it with the same environment pointer as the
procedure object being applied,thus E1 as we did before,
Slide 12.4.18
There are no parameters to bind in this frame,
Slide 12.4.19
So evaluating (ca) with respect to the global environment
reduces to evaluating the body,(set! n (+ n 1))
with respect to this new frame E3,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.20
We know what should happen in this case,We just did this!
The evaluation will again mutate the value of n to be one more
than its current value,thus changing the value from 1 to 2 in
this frame,
Slide 12.4.21
And having evaluated that part of the body,we now take the
second part,n,and evaluate that with respect to E3,Chasing up
the frame pointers finds the binding in E1,and returns that
value,2,Since this is the value of the last expression in the
body of ca,this is the value returned by the whole thing,
Thus we see how this local piece of state allows us to have a
procedure,which when evaluated in successive turns,returns a
different value,
Slide 12.4.22
So our environment model helps us understand how one
counter,ca,can have some local state which it can keep
mutating,Thus,it can return a different value each time it is
applied,
What happens when we call make-counter again,
starting from 0,but giving it a different name,
Slide 12.4.23
And just to recap,here is the environment structure we have to
this point,There is a variable ca pointing to the procedure
shown,with local state of n equal to 2,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.24
Since we want to apply make-counter to 0,we first
look up the value of make-counter with respect to the
global environment,Since it is a procedure,which is being to
applied to 0 our four-step rule applies,Step one,drop a frame,
Slide 12.4.25
Step two,convert it into an environment by scoping that frame
with the same environment pointer as the procedure being
applied,Note that this means in this case pointing to the global
environment,
Slide 12.4.26
Step three,bind the formal parameter of this procedure,n,to
the value passed in,0,Note that this is a different n than the
one we had before,This n lives in environment E4,The n we
had for ca lives in environment E1,so we have different
bindings for the same name in different environments,
Slide 12.4.27
Relative to this environment,evaluate the body of the
procedure,Remember the procedure we are applying,make-
counter,so we are going to evaluate that lambda
expression with respect to E4,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.28
And we know what evaluating a lambda should do,It
creates a procedure object,a double bubble,the code of which
has no parameter and a body that is the same as the earlier body
we created for ca,The key point is,where does the procedure
object's environment pointer point to? To the environment in
which the lambda is being evaluated,thus to E4,Note that
this procedure thus has a different scoping than the one created
for ca,It is going to capture E4,with that version of n,which
is different than the one we had before,
Slide 12.4.29
And finally,that procedure object is the value returned by
applying (make-counter 0),and therefore the
definition creates a binding for cb up in the global
environment to that new procedure object,
Look carefully at the structure we have here,We have ca
pointing to a procedure object that inherits a frame E1,We have
cb pointing to a similar procedure object that inherits a frame
E4,They have different bindings for the parameter n,and that
is going to allow these things to behave differently,
Slide 12.4.30
So lets finish our understanding of this process by looking at
what happens when we evaluate (cb)> within the global
environment,
Slide 12.4.31
Here,again,is the environment structure we have so far,We
have a global environment with a binding for ca to one
procedure object with its own frame,and a binding for cb to
another procedure object with its own frame,What happens
when we apply cb?
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.32
Well,the rules make it very clear,We take the procedure object
associated with the binding for cb,and we apply it,This will
drop a frame,scope it by using the same environment pointer as
the procedure object,which says E5 will be scoped by E4,Not
E1,not global environment,but E4,
Slide 12.4.33
This says that using exactly the same reasoning as before,the
value of n that will get mutated is the one that is seen from E5,
namely the one sitting in E4,So it gets changed by 1,and then
we return that value,of 1,
Slide 12.4.34
This says the value of the overall expression is just 1,
Slide 12.4.35
So what was the point of this long,drawn-out,exercise? Part of
it was to let you see how the rules for the environment model
evolve,They specify what happens during a computation,they
specify how expressions get meanings assigned to them through
a very mechanistic set of rules,
But the second point was to let us understand things that
involve mutation,especially mutation associated with
procedures,So think about what happened with our little
counter example,Here is the environment model we had,
6.001 Structure and Interpretation of Computer Programs,Copyright? 2004 by Massachusetts Institute of Technology,
Slide 12.4.36
And notice that there is a structure associated with it,What does
ca look like? It's a name,bound in the global environment,
that points to a procedure that has as its environment pointer,a
pointer into a frame,in this case E1,that has some local state in
it,
The only way we can get at that value of n is through a
procedure whose environment pointer,through some chain of
frames,points into that frame,In this case,that is only through
ca,
So this structure or pattern is a common pattern,
Slide 12.4.37
And in fact when we evaluated make-counter a second
time and gave the result returned by it the name cb,we created
a similar pattern,but now with its own local state,Thus,each of
these names specifies a procedure that has associated with it
some information that belongs only to that procedure,
This kind of framework,that is,procedures that capture local
state and can manipulate local state,is going to be a very useful
programming tool,as we are going to see in the next few
lectures,
Slide 12.4.38
Here is a summary of the key points to take away from this
exercise,
