Mod-02 Lec-03 Run-time Environments-Part 2

Welcome to part 2 of this lecture on run-time
environments. In the last lecture, we discussed some aspects of run-time support; for example,
what is run-time support, parameter passing mechanisms and a little bit of storage location. Today, we will continue the part 2 of the
lecture with the topics that we have not yet discussed. A little bit of review from last
time: we discussed static data storage allocation. Basically, the compiler is responsible for
allocating space for all variables both local and global in this type of allocation.
This is done for all procedures at compile time. If you look at the picture here, you
have main program variables, procedure P 1 variables, P 2, P 3 variables and so on and
so forth. The difficulty with this allocation method
is we cannot have recursion. No stack and heap allocation is available here. There are
no overheads, but at the same time we cannot have recursion. Dynamic storage allocation is another policy,
where compiler locates place only for global variables at compile time, but space for all
other variables of procedures will be allocated only at runtime. This policy is based on stack
heap allocation and it is available in C, C plus plus, Java, FORTRAN 8 and FORTRAN 9.
The variable access in this case is a bit slow because there are several indirections
in accessing locations, but recursion can be implemented here. This is the biggest advantage
in dynamic storage location. We also saw the activation record structure.
An activation record is a unit of allocation for the run-time support. Whenever a procedure
is called, a new activation record is created; parts of the activation record are initialized
by the caller and parts of the activation record are initialized by the callee. There
is almost everything that you need, all the information: return address, some static and
dynamic links, address of the function result and then the actual parameters, local variables,
temporaries, machine status, that is, the set of registers and other information, space
for local arrays, etcetera. The first thing that a compiler should do
during intermediate code generation itself is to compute the offset at which variable
is going to be stored in the activation record. If the storage allocation is static, then
these are going to be offsets in the static data area itself, but otherwise these are
all offsets in the activation record. The compiler should compute the offsets at which
variables and constant will be stored in the activation record.
These offsets will be with respect to the pointer pointing to the beginning of the activation
record and variables are usually stored in the activation record in the declaration order
itself. For example, if you have int A, B, C semicolon; float D, F something like that,
then the offsets for A, B, C will be 2, 0, 4 and 8; and the next D and F will get offset
from 8 onwards that is 12 and 12 plus 8; that is 20. It is in the storage in on declaration
order itself. We will see a little more detail of how is this computation is performed during
semantic analysis. For example, consider a simple grammar program
consisting of only declarations. P going to decl and a declaration is type T followed
by id semicolon Decl 1; Decl 1 is the other non-terminal which produces more and more
declarations. Otherwise the ending terminal production is decl going to T type and followed
by id. Type can be int, float or num as indicated
in the other 3 productions here. Let us start looking at the method by which the variable
offset is computed. The declaration is the most important non terminal. It has actually
an inherited attribute called inoffset and it also has a synthesize attribute called
outoffset. To begin with, at the beginning of the declaration
list this inoffset is set to 0 and then as we parse declarations, for example, look at
this now T id semicolon declaration 1. This identifier is entered into the symbol table
along with the name, the type and the inoffset that is the beginning of that declaration
list there is some offset which comes in to this list. That is the offset which is assigned
to the name id. Now the declaration 1 dot inoffset is also set as declaration dot inoffset,
but it is added an increment of T dot size. The reason is this declaration id is of type
T and T dot size is the size of that particular type. For example, ints may take 4 bytes;
floats may take 8 bytes and so on and so forth. That is why, whatever we got from the left
hand side declaration non terminal is incremented by the size of that particular type and that
is assigned to declaration 1 dot inoffset. What is declaration dot outoffset? That is
whatever we get from declaration 1 dot outoffset. The declaration which are parsed by this non
terminal or generated by this non terminal are all accumulated and then outoffset will
give you the final offset for this particular thing.
This recursive production applies itself again and again and generates all the declarations.
Each time a declaration is produced there is some incrementing and then the next declaration
gets that added offset. For example, declaration going to T followed by id terminates this
recursion and at that point we enter the name with type and some offset and outoffset is
T dot size itself because this is a declaration which is terminating the entire thing.
This is how we parse the declarations and assign variable offsets. T going to int – size
is 4, T going to float – size is 8 and T going to num T 1 is an array. The T dot type is
array type and T dot size is T 1 dot size into the num dot value that is the size of
that particular array is num dot val. This num dot value is multiplied by the type of
this particular array whose size is available in T1 dot size. This is an example of how
variable offset computation is performed in the declaration order. Now we come up to a very important part of
this lecture, how are activation records allocated during dynamic allocation that happens in
languages such as C and C plus plus. Let us consider a simple program here, program RTST
and the procedure is P, which is actually enclosed within the main program RTST. Then
procedure Q is enclosed within procedure P and the procedure Q has a body in which we
say begin, call the procedure R and end; procedure R is just outside procedure Q; it also has
a body which says begin, call the procedure Q and end.
So, basically Q calls R and R calls Q there is some recursion which happens here and at
the level of procedure P itself, there is a body which says call R and the main program
begins by calling P. Here is a call sequence which is possible. The main program RTST calls
P and P calls R, R calls Q and Q calls R. We are going to trace the allocation of activation
records for this particular sequence. What we need to remember at this point is activation
records are created at procedure entry time and they are destroyed at procedure exit time.
To begin with, what we have is the main program space in the stack of activation records.
In the stack heap area we just have the main program data already available because these
are like global variables. Whatever is stored here is a set of global variables which can
be accessed by any of these procedures; that is the program variables which are declared
here. There is a static link chain which will be
shown in the following diagram on this side and there is a dynamic link chain which will
be shown in the diagram on the other side. So, let see. Next is the next place where
the activation record can be created in the stack heap. So, RTST has called P, the activation record
for P did not exist; it is created now. Now there are 2 questions to be answered: One
is what to do with the dynamic link chain and the other one is what to do with the static
link chain. The rest of the initialization within the activation record is fairly straight
forward and there is nothing much involved there; various return addresses and temporaries
etcetera are all stored there. The dynamic link chain is used only to maintain
the stack nature of this particular set of activation records. As we go on, as we add
1 activation record at a time, this dynamic link will point to the previous activation
record. When we remove for example, this particular activation record, the activation for P ends
after the call chain comes back. This entire thing will be removed; this particular link
also disappears. What is the static link? The static link actually
allows us to access the global variables. For example, if you look at the scope of the
variables which are visible within the procedure P, here is the body of the procedure P, this
particular body can access the variables of program RTST; they can also access the variables
of the procedure P. As far as the variables of program RTST are
concerned, they are like global variables and the variables within the procedure P are
the local variables. The access to local variables is very easily done with respect to a pointer
call the base pointer. All the offsets, we saw offset computation example before, offsets
are all with respect to this base pointer. So, it is very easy to add the offset to the
base pointer and get to the location where the local data is stored.
For each variable, we have an offset and adding that to this base will give us the location
where the variable is stored. Suppose we have global variables that is the variables of
the main program RTST which we need to access within the procedure P, in such a case, the
static link chain is going to help us. Take the static link, see where it points, it is
actually pointing to RTST itself. If we take the contents of this particular
static link variable, we get the base pointer of this particular RTST that is the way it
would be. Once we know the base pointer of RTST, accessing the variable within that is
really straight forward. Again we have the offset that has already been computed and
it is easy to access it. All this happens at compile time. In other
words, the compiler already knows because of the nesting levels of the procedures, how
many times it needs to skip the static link. This will become clear as we go along and
then treat that as a pointer to the activation record and access the variables within that
activation record. Let us now look at the call to R from P. See
the activation record for C is now created fresh. The dynamic link simply points to the
previous one, previous activation record that is the tough P and then whatever remains as
dynamic link from P to RTST remains. This is the chain of dynamic links used to
maintain the stack nature of the activation records. On this side look at the procedure
R; the procedure R is just within the procedure P. So, the body of the procedure R can access
the variables which are within the procedure R, it can access the variables of the procedure
P which is the enclosing procedure and it can also access the variables of RTST which
is the next level enclosing procedure – that is the main program itself.
Using the base pointer, the code can access all the variables of R by using one level
of indirection on the static link and then treating that as the base pointer it can access
all the variables of P. By doing indirection twice, it can get the base pointer of the
main program RTST and access the variables of RTST.
Remember that the compiler itself knows how many times this indirection needs to be applied
because the symbol table will contain the nesting depth of each one of the program variables
and the procedures. We go one step further, R calls Q. We had
RTST which called P; then P called R; now R calls Q. R and Q happen to be at the same
nesting level. They are not nested within one another.
If you look at Q, the body of Q can access the variables within procedure Q, it can access
variables within the procedure P and it can access the variables within the main program
RTST. That is all, just like the procedure R. It is not allowed to access the variables
of procedure R. So, making the static link point blindly to the activation record of
R will be incorrect. If that is done, then the variables of R can also be accessed which
is not right. Therefore, this particular static link should
be made to point to the beginning of the activation record of P itself. Now, whenever the compiler
needs, it can generate code to access the variables of this procedure Q. By doing one
indirection, it can access the variables of P and by doing 2 indirections, it can access
the variables of RTST. The variables of R will not be accessed at all, which is the
right thing to do. Let us look at the next activation which is
that of R. So far, we had RTST then called P, P calls R, R calls Q and then there is
a recursion, Q calls R again. There are 2 instances of R and an instance of Q. So, R
cannot and should not access the variables of Q – this R, the second instance of R and
it should not also access the variables of the first instance of R. It should actually
be accessing only the variables of R, P and RTST and this is appropriately done by making
the static pointer or the static link of this particular activation record point to the
beginning of the activation record of P. The next thing that we need to see is how exactly
are we going to establish these static links. Dynamic links are very easy. This is dependent
on the level information of the various procedures. For example, RTST is assumed to be at level
1; procedure P is assumed be at level 2; then procedure Q is at level 3; procedure R is
also at level 3. The levels of the procedures are all indicated in this caller chain. RTST
has superscript 1, P has 2, R has 3, Q has 3 and R has 3. Let us see how it is established. The basic
rule is very simple: skip L1 minus L2 plus 1 records starting from the caller’s activation
record and establish the static link to the activation record reached.
What is L1? L1 is the level of the caller and L2 is the level of the callee. Let us
take this call chain and look at an example, 2 examples really. Consider the call P to
R; that means this RTST would already be present; P will already be present, but R is not yet
present; of course, these 2 are not yet present. So, the level difference is 2 minus 3 plus
1 which is 0. Hence, the static link of R should point to this activation record of
P itself. That is the simplest rule. Similarly, take R calls Q; we should really be pointing
here. Let us see whether we do that. So, 3 minus 3 plus 1 is 1. Hence, skipping
one link starting from R, we get P and the static link of Q will point to the activation
record of P. This is how the static link chain is set and this formula, which is very simple
one, enables us to set this links appropriately. What we saw so far is one method of arranging
activation records; this is using the static link and dynamic link. There is another way
of establishing access to these global variables and arranging the activation records; that
is known as the display stack. Let us discuss this display stack and then
see what is the advantage or otherwise of each of these schemes. Take the same program
RTST, P, Q and R. The display stack is a very simple stack of pointers to the activation
records. To begin with the display stack has a pointer to the activation record of RTST,
the main program and nothing else. Activation records are going to be stored exactly the
way they were stored before – that is in the stack heap area.
The creation of activation record, destruction of activation record etcetera happens as before;
there is no difference at all. To begin with, we have just this RTST; one pointer pointing
to the activation record of RTST. Now there is a call; the call is to the procedure
P. When there is a call, we actually push the activation record pointer of the callee
or we decide to pop some of the entries of the display stack and then push the activation
record pointer; one of these can happen based on this particular formula.
The formula says: Pop L1 minus L2 plus 1 records off the display of the caller and push the
pointer to the activation record of callee on to the stack. L1 is the caller, L2 is the
callee and the formula is exactly the same as before.
Let us do that here. The levels, when we look at it, RTST is at level 1, P is at level 2;
1 minus 2 plus 1 is 0. So, nothing needs to be popped from this display stack; you only
push the pointer to P’s activation record. Then P calls R; P is at level 2 and R is at
level 3. 2 minus 3 is minus 1 and then plus 1, again 0. So, we again need not pop anything
from this display. You only push the activation record pointer of R on to this display stack,
but now something different happens. R now calls Q, both are at level 3. 3 minus 3 is
0 plus 1 is 1. So, we need to pop one activation record pointer;
that is that of R from this stack which uncovers the pointer P and then we push the new pointer
Q on to the stack exactly the way we did before. There is no difference at all. There, the
static link changes allowed us to access the various global variables. Here, as I am going to tell you very soon,
the stack of activation record pointers – the display stack allows us to do exactly the
same thing. We get 3 activation record pointers Q, P and RTST on this and then once this happens,
there is another call to R again; 3 minus 3 plus 1 is 1. So, one of the pointers that
is Q is pushed out R. The new R activation record pointer is pushed on to the stack and
from now onwards the recursion unwinds. This is where this particular comment becomes important.
The popped pointers are stored in the activation record of the caller and restored to the display
after the callee returns. In other words, whatever we popped here cannot be just dispensed
with. We need to store it in the activation record of the caller and then once we return,
this particular pointer will have to be pushed on to the stack again and situation should
be exactly as before. This is after the callee returns; that is
what we do. The callee will not know about this activation record pointers at all. See
for example, when R calls Q we have dispensed with the activation record of R, rather the
pointer to activation record of R is removed from here and then the pointer to the AR of
Q is inserted. Therefore, the callee cannot store this pointer in its memory, only the
caller will have to store the pointers which are going to be popped off and then place
the call to the new procedure. Let us see how the variables are accessed
within these programs. Let us take this example. We have Q, we have P, we have RTST. These
are the 3 activation records which are active at this point of time and the code of procedure
Q can access the variables of itself, those of P and those of RTST; that is what this
really means. This can be done very easily here. Q is a
pointer to the activation record; this is the pointer to the Qs activation record. Using
this pointer, it can access the AR of Q and variables of Q with an offset.
Similarly at the next level, if there is a variable of P which is easily known from the
symbol table, the compiler generates code to take the pointer AR which is mentioned
here as P, go to that particular activation record, use the offset and dig into the AR
of P. Similarly for RTST also. The access is very similar to what happened
before. Only thing is we really do not have to traverse any static link chains. Here,
we just use these pointers which are on the display stack in order to access the variables. Before we begin this static scope and dynamic
scope, let me mention a few points about the display stack and the static link, dynamic
link chain itself – the methods themselves. When you have the sldl scheme, you need to
traverse the static link chains possibly many times before we access the right global variables.
There are many indirections which can happen. This can become a bit slow if the number of
global variables in the program is large. This does not happen in the case of the display
stack. The pointers are already on the stack and taking them from the stack is no overhead,
but popping and pushing the pointers from the display stack is extra in the case of
the display stack mechanism. This does not happen in the case of the sldl scheme.
It is very difficult to say which particular method is more efficient compared to the other
and there are compilers which use either of these methods. We are not going to get in
to a debate on which is better; both are equally good and any one can be used
Let us proceed further; the next concept that we discuss now relates to the scope. There
are 2 varieties of scope: first is the static scope and the second is the dynamic scope.
Let us understand these words well. When the scope is said to be static, a global identifier
or a name, global name refers to that particular identifier with that name that is declared
in the closest enclosing scope of the program text. It uses the static relationship between
blocks in the program text. That is, this is exactly the way we understand
the scope of variables in C and C plus plus. Look at the program, see at which level the
variable is declared and whenever we want to refer to a variable, we just go up the
nesting levels, find the first place where it is declared and that is our variable.
That is, what we mean by saying the name refers to the identifier with that name that is declared
in the closest enclosing scope of the program text. You just go up the nesting levels of
for the procedures, find the first one which contains that particular name. This particular
relationship does not change. It is related to the program text and it is not related
to the execution of the program but dynamic scope is different.
We are going to see example of this very soon. A global identifier refers to the identifier
associated with the most recent activation record. For example, if we have 2 procedures
with the same variable x depending on which is active a point, x may refer to either one
of them. We will see an example of this. It uses the actual sequence of calls that are
executed in the dynamic execution of the program and both are identical as far as local variables
are concerned; there is no different at all. Let us see a simple example. Here is a C like
program: int x equal to 1; function g z, the body is x plus z, it is just an expression;
function f y has a bigger body, it has a variable x equal and it is assigned the value y plus
one and it returns g of y star x; this particular function f is called with f 3 in the main
program. Let us assume static scope. What really happens
is, when we are within this function g z the x here, there is no declaration of x within
the function g; it just goes to the next nesting level that is, the main program. This particular
x is what we refer here; that is static scope. When we come to a function f y, you have int
x equal to y plus 1; this is a local variable which is declared within the function f. Then
we say return g of y star x. We have called f with 3. So, y is 3, x becomes 4 and g is
called with 4 into 3 – that is 12. When we come to g z, let us trace what happens.
This says x plus z, z is already 12, but the x we refer to is 1. So, 12 plus 1, 13 is returned
as the value of f. This is the way a program executes when it is using static scope.
Most importantly, even though the function f y is active, when g is executing; for example,
we called f. See, you look at this particular chain This is the outer block x with value
one; this is the activation record corresponding to f of 3; the variable y and variable x are
present here; y is the parameter, x is the local variable; y has the value 3, x has the
value 4 and then we have another activation record corresponding to g where the value
of z, which is the parameter, is 12. Let us see what happens if we use dynamic
scope. Let us trace the whole thing all over again we called f with 3; x becomes 3 plus
1 4; then g is called with y star x that is g of 12. So far so good, there is no change
with what happened. But when the function g is called it says
x plus z. In dynamic scope, the question asked is which x? In static scope, it was very easily
resolved as this particular outer x, the enclosing main program is looked at and x is determined
to be the variable which is relevant here. In dynamic scope, it looks at the most recent
activation of some function which contains the variable declaration for x. In this case,
it is this particular f. So, that is why it is so. Let me go back.
We have g here, the parameter is 12, but when we come here x plus z, the most recent activation
of x corresponds to the variable int x within the function f y here. This x is not this
x which is relevant, when dynamic scope is active. In static scope, this is always the
variable that is relevant whereas, in dynamic scope this is the most recent activation of
x; so this x is relevant. What we do is add 4 to 12 and return 16. In
the previous case, we had added one to 12 and returned 13. Now, we return 16. This is
the precisely the way the dynamic scope works. Here is another example. Let us see what happens.
Here is the floating variable r, 0.25; then the procedure show simply prints the value
of r; the procedure small has a local variable called r which is initialized to 0.125 and
then it calls show. What does the main program do? The main program
calls show and then calls small and prints a new line. It does the same thing all over
again. Let us trace the values which are printed out by these 2 print statements under static
scope and dynamic scope. Under static scoping, show prints r; the only r which is relevant
at this point is the r here, which is printed out as 0.25. Then it calls small. It has a
float variable called r which is initialized to 0.125, but when we call show under static
scope, again there is no other r except this which is relevant. So, 0.25 is printed all
over again. The same goes for the next line — 0.25 and 0.25.
Let’s see what happens under dynamic scope. The first show – there is only one r, which
is active that is this r; so, it prints 0.25. Then it calls small. Now, the second r becomes
active because the procedure small is called. The new r has 0.125. When show is called,
the r variable that is referred to here is the most recent activation of r which is this.
So, it prints 0.125. When it comes out and prints a new line and then show is called
again, at this point this procedure small and the variable r have vanished; the activation
record have already been destroyed because the procedure has returned. 0.25 is printed
all over again. small repeats exactly what happened. The small creates a new variable,
it calls show and this r within show is again going to refer to the r which is inside small.
0.125 is printed all over again. So, this is how the static scope and dynamic scope
really work Let see how to implement the dynamic scope.
Static scope, we already saw how to implement it; that was using the static link dynamic
link chain or using the display stack. Here we have what is known as the deep access method
and in the next few minutes after this, we will also discuss another method call the
shallow access method. What is the deep access method? In this case,
we use the dynamic link itself as the static link. In other words, we do not have 2 links;
we just have 1 link. It is dynamic link Static link is not necessary. The reason is simple;
we always want the most recent activation of the variable and we are not worried about
the scope. So, there is no need to compute the value of static link as we used to do
before. Now we just use the dynamic link, but how to find the most recent activation
of that particular name. This is done by searching the activation records on the stack; to find
the first activation record containing the non local name. If it is a local name, then
it is within my activation record. Whatever I am doing within my procedure activation,
the variable is existing. If it is a global variable, we need to go
out, but how far into the stack of activation records is the question. Since we do not know
how far, it is called as the deep access method. It can go up to the main program, that is,
the first activation record or it could be the immediate previous predecessor activation
record itself; we have no idea. The depth of the search depends on the input
to the program and cannot be determined at compile time. This search also needs some
information on the identifiers to be maintained at runtime within the activation records.
Why? How do we search for a variable in an activation
record? If the activation record has only offsets, there is no way we can say, this
is that of the variable r or x or a. So, for each variable, we also need to maintain some
type of an integer code saying with this code, this is the variable associated; with this
code, this is the variable associated and so on and so forth.
Once we have this coding within the activation record, that is, the variable and the integer
code which is already known to the compiler, this can be established by the compiler; that
is not a problem. Because each name needs to have a unique code nothing more than that
is necessary. We search for the existence of that particular code. Once that code is
available, the offset corresponding to it is used for access to that variable.
Such a mechanism takes much longer time to access the globals, but there is no overhead
when activations begin and end. This is an advantage. Activations do not cause any overheads,
but when you want to access globals, you have to get deep into the stack of activation records
and then access the variable. Let us look at the next method of implementing
dynamic scope. This is called as the shallow access method. Why is it called shallow? It
will be very clear soon. It is a very simple method, allocates some storage for each name.
In other words, if the name has 5 or 6 instances that is, the same name is declared in many
procedures it does not matter. We just want to look at each name and allocate
the maximum amount of storage for that particular name. In other words, let us say the variable
is x; it may be declared as int in a particular function or procedure. In another function
or procedure, it may be declared as an array; and in third one it could be a struct. What
we want is the maximum storage corresponding to this name; it could be the array, it could
be the struct, any one of these possible. Of course it can be a single variable, but
it can be an int but struct is not smaller than int; that is, why I said either struct
or the array itself. So much maximum storage is known already and
that is the storage which is allocated statically for each name. There is a new unique location
in each name, very simple. When a new activation record is created for a procedure p, a local
name n in p takes over the static storage allocated to the name n. Therefore, there
is only 1 location or set of locations corresponding to the variable n.
Whenever there is a need for that variable n, it could be in a particular activation
of a procedure, we just use this static region as the storage allocated to the name n. The
previous value of n is held in static storage As we go along there could be another n which
comes along. Then the previous n must vacate the space and give it to the new n. What happens
to the value of old n? The previous value of n held in static storage
is saved in the activation record of p and restored when the activation of p ends. Now
it is the callee which is storing the variable. Before it tries to access the storage for
n, whatever is already available in n is stored in its own activation record. We know the
maximum size of that variable storage. That is why we can store it in the activation record.
Once the activation of p ends, whatever is in the activation record, the previous value
is restored into the static area and then the call continues.
Direct and quick access to globals is possible because this is like hashing. You always come
to one location for n; it is static, the address is known, but some overhead is incurred when
activations begin and end. You need to store the old value of n and restore it when the
procedure p ends. So, this is the overhead that is incurred. Whereas, in the previous
case that is, the deep access method there was no overhead when activation begin and
end, but you have to traverse deep in to the stack in order to access that particular variable. The next topic that is very important is to
understand how to pass functions to other functions as parameters. To understand this,
we must understand what exactly first class functions are. Here is a description:
A language has first class functions if functions can be declared within any scope. For example,
Pascal has such a facility and functions can be declared within classes in C plus plus
that is, the methods and functions can be passed as arguments to other functions. This
is something we have not discussed so far and they can also be returned as results of
functions. When we say passed as arguments, we are not looking at a function call as a
parameter. It will become very clear with this particular example
Let us look at the example and then return to the text again. Here is main; you have
int x equal to 4 and then you have int f which is a function f which returns an integer value,
it takes another integer as a parameter and returns x star y, so far so good.
There is another function called g which takes a function as a parameter. The function parameter
is h and the function takes as input an integer and returns output as another integer. That
is why this is a mapping from int to int. This is the type of this particular h. Just
like int was the type of the parameter y here, the type of the parameter here is a map from
int to int. The body contain int x equal to 7, straight
forward and then it is says return h 3 plus x which is straight forward again. Let us
start looking at the call to g with a parameter f. Notice that, this f is not loaded with
any parameters at this point. g has simply an actual parameter called f which is nothing,
but this function which is passed as a parameter. Let us go to g. Types match here; f is int
to int, takes input as int and returns output as int. So, int to int is matched here. Now
x is initialized to 7. When you say return h 3 plus x that f corresponds to h.
It is really a call to f with the parameter 3. Now, the function f is executed. At this
point when we passed f to g, there was no execution of the function f, but here when
we call the function h which is nothing, but the function f along with the parameter 3,
the actual function f is invoked. That is x star y; it returns x star y. We have 3 and
then we have 4 here; so, the return value is 12, plus x again. This 3 star y; g of f
returns h of 3; int y was 3; 3 into 4 was 12. This would not return 12. There is a minor
mistake here. So, 12 plus 4, 16 is returned here. This should have been 16.
In a language with first class functions and static scope, a function value is generally
represented by a closure. We know how to pass integers, floats and so on and so forth as
parameters. Now, let us learn how to pass functions as
parameters. It is not enough to pass a pointer to the code itself. We need to pass a pointer
to the function code and also a pointer to an activation record. Which activation record?
It will become very clear with an example in the next slide.
Passing functions as arguments is very useful in structuring of systems using what are known
as up calls. For example, you may have a nested set of procedures. One nested in another and
the inner most procedure probably is required to be called at the outer most level which
is not possible if you look at static scoping. In order to permit such a thing to happen
even under static scope, we can pass the inner most function as a parameter and then execute
it whenever it is necessary in the outer function. This is called an up call and it is very useful
in operating systems design and implementation. Here is an example of closure. Let us explain,
what exactly is a closure here. Here is the example that we had seen earlier. You have
the main program that is, the variable x equal to 4 and nothing else, only one variable.
When we call g of f within the main program, here is the activation record corresponding
to that particular call to g; there is a static link which points to the main program. We
are looking at static scope. The only things that we can see here from within g are the
variables of g itself and the variables of main. Nothing else can be seen that is, we
can see x and f that is all. Now here is the integer variable within the
function g and the parameter which was passed as f to g. This is the closure corresponding
to the parameter h: one part is a pointer to code for f and a second part is a pointer
to the enclosing scope that is, the tough main itself.
When the call to h that is, the call to f is made, the only difference in the implementation
is creation of the activation record remains the same, but creation of the static link
is actually done by copying this particular static link from the closure parameter to
the actual static link variable and there by pointing it to main itself.
This particular activation of f runs as if it is an activation with access to itself
and also the activation record of main. This is the only difference. Let me repeat when
we create the activation record of the function parameter, the static link is created by copying
the static link value from the closure parameter and it is not computed using L1 minus L2 plus
1, but when we pass a closure which particular static link to be passed is computed using
the same formula L 1 minus L 2 plus one treating f, as if it were a call.
f is at this level. This is let us say 1; this is 2. If you treat it as a call to f,
you really have 1 minus 2 plus 1 which is 0, that is, the pointer to the activation
record of main itself is to be passed to this particular closure. That is how we determine;
it is the same L1 minus L2 plus one. Whatever I explained so far is listed here.
Just to make sure that we understand everything well let us go through it very quickly. In
this example when executing the call h 3 h is really f and 3 is the parameter y of f.
This parameter y is nothing, but this particular 3. Without passing a closure, the activation
record of the main program cannot be accessed. How do we know what exactly is the scope of
this particular h? Hence the value of x within f will not be 4. We have no idea what this
particular x will be because we do not even know that we can access the main program.
When f is passed as a parameter in the call g f, a closure consisting of a pointer to
the code for f and a pointer to the activation record of main program is passed. This is
what I mentioned now. When processing the call h 3, after setting up an activation record
for h, the static link for the activation record is set up using the activation record
pointer in the closure for f that has been passed to the call g f. This was what I was saying just now. We really
copy this to this, rather than computing this all over again using L1 minus L2 plus one.
That is really as far as the activation record management is concerned. That is how exactly
static scope, dynamic scope etcetera are taken care of. Now let us switch to a slightly different
topic called as the heap memory management. What is heap? Heap is just an area in memory.
There is a difference between heap and stack. Stack is a very systematic data structure;
there is only push and there is only a pop permitted and a stack can be implemented using
either link layer star or it can be implemented using an array.
A heap is extremely useful for allocating space for objects created at run time. We
do not know when the object is going to be created. Therefore, it is not possible to
use the stack; you can only use the heap. For example, nodes of dynamic data structure
such as linked list and trees cannot be created using a stack or an array etcetera without
losing some flexibility. Dynamic memory allocation and deallocation
based on the requirements of the program. For example, C programs have malloc. They
do malloc when they want a node to be created in a link list or in a tree etcetera. They
call free when that node is not needed anymore. Similarly, in C plus plus programs, there
is a call to new to create a new object and then there is call to delete the old object.
In Java, there is only new that is, a call to new is going to create a new object, but
there is no call to delete or free any object. There is a process called garbage collection
which collects all the useless objects and returns it to some pool.
Allocation and de-allocation may be completely manual as in C, C plus plus. That is we have
malloc, free, new and delete; semiautomatic as in Java that is, we have new and then garbage
collection or it could be fully automatic as in the case of a functional programming
language such as Lisp. We will actually end our lecture at this point
and in the next class we will discuss more details of how heap is organized. We are going
to look at a memory a manager, how the allocation happens, how de-allocation happens, etcetera.
Thank you very much.

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  • Vikrant Singh says:


  • Suyash Singh says:

    Can you please share the link of part 1

  • AgentFriday says:

    at 19:02…  This is completely wrong.  Skipping "L1-L2 +1 activation records", if you carry it to the first Q -> R call, you would skip only (3-3) +1 activation records, which points from the 2nd activation record of R to the first.  What good does that do?  This needs closer examination and explanation.

  • RISHU ROSHAN says:

    How u calculate offset of T id; decl1…
    T Id and decl1 are different ,how u calculate offset of decl1 by using size of T??? They are different pr

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