bp and sp registers
to access and allocate such variables. Consider the following Pascal program: program LocalStorage;
var i,j,k:integer;
c: array [0..20000] of integer;
procedure Proc1;
var a:array [0..30000] of integer;
i:integer;
begin
{Code that manipulates a and i}
end;
procedure Proc2;
var b:array [0..20000] of integer;
i:integer;
begin
{Code that manipulates b and i}
end;
begin
{main program that manipulates i,j,k, and c}
end.
Pascal normally allocates global variables in the data segment and local
variables in the stack segment. Therefore, the program above allocates 50,002
words of local storage (30,001 words in Proc1 and 20,001 words
in Proc2). This is above and beyond the other data on the stack
(like return addresses). Since 50,002 words of storage consumes 100,004
bytes of storage you have a small problem - the 80x86 CPUs in real mode
limit the stack segment to 65,536 bytes. Pascal avoids this problem by dynamically
allocating local storage upon entering a procedure and deallocating local
storage upon return. Unless Proc1 and Proc2 are
both active (which can only occur if Proc1 calls Proc2
or vice versa), there is sufficient storage for this program. You don't
need the 30,001 words for Proc1 and the 20,001 words for Proc2
at the same time. So Proc1 allocates and uses 60,002 bytes
of storage, then deallocates this storage and returns (freeing up the 60,002
bytes). Next, Proc2 allocates 40,002 bytes of storage, uses
them, deallocates them, and returns to its caller. Note that Proc1
and Proc2 share many of the same memory locations. However,
they do so at different times. As long as these variables are temporaries
whose values you needn't save from one invocation of the procedure to another,
this form of local storage allocation works great. procedure LocalStuff(i,j,k:integer);
var l,m,n:integer; {local variables}
begin
l := i+2;
j := l*k+j;
n := j-l;
m := l+j+n;
end;
Calling sequence: LocalStuff(1,2,3);
Assembly language code:
LStuff_i equ 8[bp]
LStuff_j equ 6[bp]
LStuff_k equ 4[bp]
LStuff_l equ -4[bp]
LStuff_m equ -6[bp]
LStuff_n equ -8[bp]
LocalStuff proc near
push bp
mov bp, sp
push ax
sub sp, 6 ;Allocate local variables.
L0: mov ax, LStuff_i
add ax, 2
mov LStuff_l, ax
mov ax, LStuff_l
mul LStuff_k
add ax, LStuff_j
mov LStuff_j, ax
sub ax, LStuff_l ;AX already contains j
mov LStuff_n, ax
add ax, LStuff_l ;AX already contains n
add ax, LStuff_j
mov LStuff_m, ax
add sp, 6 ;Deallocate local storage
pop ax
pop bp
ret 6
LocalStuff endp
The sub sp, 6 instruction makes room for three words on the
stack. You can allocate l, m, and n in these three
words. You can reference these variables by indexing off the bp
register using negative offsets (see the code above). Upon reaching the
statement at label L0, the stack looks something like:This code uses the matching add sp, 6 instruction at the
end of the procedure to deallocate the local storage. The value you add
to the stack pointer must exactly match the value you subtract when allocating
this storage. If these two values don't match, the stack pointer upon entry
to the routine will not match the stack pointer upon exit; this is like
pushing or popping too many items inside the procedure.
Unlike parameters, that have a fixed offset in the activation record, you
can allocate local variables in any order. As long as you are consistent
with your location assignments, you can allocate them in any way you choose.
Keep in mind, however, that the 80x86 supports two forms of the disp[bp]
addressing mode. It uses a one byte displacement when it is in the
range -128..+127. It uses a two byte displacement for values in the range
-32,768..+32,767. Therefore, you should place all primitive data types and
other small structures close to the base pointer, so you can use single
byte displacements. You should place large arrays and other data structures
below the smaller variables on the stack.
Most of the time you don't need to worry about allocating local variables
on the stack. Most programs don't require more than 64K of storage. The
CPU processes global variables faster than local variables. There are two
situations where allocating local variables as globals in the data segment
is not practical: when interfacing assembly language to HLLs like Pascal,
and when writing recursive code. When interfacing to Pascal, your assembly
language code may not have a data segment it can use, recursion often requires
multiple instances of the same local variable.
Recursive proc
call Recursive
ret
Recursive endp
Of course, the CPU will never execute the ret instruction at the end of
this procedure. Upon entry into Recursive, this procedure will immediately
call itself again and control will never pass to the ret instruction. In
this particular case, run away recursion results in an infinite loop.Recursive proc
jmp Recursive
ret
Recursive endp
There is, however, one major difference between these two implementations.
The former version of Recursive pushes a return address onto the stack with
each invocation of the subroutine. This does not happen in the example immediately
above (since the jmp instruction does not affect the stack). Recursive proc
dec ax
jz QuitRecurse
call Recursive
QuitRecurse: ret
Recursive endp
This modification to the routine causes Recursive to call itself the number
of times appearing in the ax register. On each call, Recursive decrements
the ax register by one and calls itself again. Eventually, Recursive decrements
ax to zero and returns. Once this happens, the CPU executes a string of
ret instructions until control returns to the original call to Recursive.
Recursive proc
RepeatAgain: dec ax
jnz RepeatAgain
ret
Recursive endp
Both examples would repeat the body of the procedure the number of times
passed in the ax register. As it turns out, there are only a few recursive
algorithms that you cannot implement in an iterative fashion. However, many
recursively implemented algorithms are more efficient than their iterative
counterparts and most of the time the recursive form of the algorithm is
much easier to understand. procedure quicksort(var a:ArrayToSort; Low,High: integer);
procedure sort(l,r: integer);
var i,j,Middle,Temp: integer;
begin
i:=l;
j:=r;
Middle:=a[(l+r) DIV 2];
repeat
while (a[i] < Middle) do i:=i+1;
while (Middle < a[j]) do j:=j-1;
if (i <= j) then begin
Temp:=a[i];
a[i]:=a[j];
a[j]:=Temp;
i:=i+1;
j:=j-1;
end;
until i>j;
if l<j then sort(l,j);
if i<r then sort(i,r);
end;
begin {quicksort};
sort(Low,High);
end;
The sort subroutine is the recursive routine in this package. Recursion
occurs at the last two if statements in the sort procedure. include stdlib.a
includelib stdlib.lib
cseg segment
assume cs:cseg, ds:cseg, ss:sseg, es:cseg
; Main program to test sorting routine
Main proc
mov ax, cs
mov ds, ax
mov es, ax
mov ax, 0
push ax
mov ax, 31
push ax
call sort
ExitPgm ;Return to DOS
Main endp
; Data to be sorted
a word 31,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16
word 15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0
; procedure sort (l,r:integer)
; Sorts array A between indices l and r
l equ 6[bp]
r equ 4[bp]
i equ -2[bp]
j equ -4[bp]
sort proc near
push bp
mov bp, sp
sub sp, 4 ;Make room for i and j.
mov ax, l ;i := l
mov i, ax
mov bx, r ;j := r
mov j, bx
; Note: This computation of the address of a[(l+r) div 2] is kind
; of strange. Rather than divide by two, then multiply by two
; (since A is a word array), this code simply clears the L.O. bit
; of BX.
add bx, l ;Middle := a[(l+r) div 2]
and bx, 0FFFEh
mov ax, a[bx] ;BX*2, because this is a word
; ; array,nullifies the "div 2"
; ; above.
;
; Repeat until i > j: Of course, I and J are in BX and SI.
lea bx, a ;Compute the address of a[i]
add bx, i ; and leave it in BX.
add bx, i
lea si, a ;Compute the address of a[j]
add si, j ; and leave it in SI.
add si, j
RptLp:
; While (a [i] < Middle) do i := i + 1;
sub bx, 2 ;We'll increment it real soon.
WhlLp1: add bx, 2
cmp ax, [bx] ;AX still contains middle
jg WhlLp1
; While (Middle < a[j]) do j := j-1
add si, 2 ;We'll decrement it in loop
WhlLp2: add si, 2
cmp ax, [si] ;AX still contains middle
jl WhlLp2 ; value.
cmp bx, si
jnle SkipIf
; Swap, if necessary
mov dx, [bx]
xchg dx, [si]
xchg dx, [bx]
add bx, 2 ;Bump by two (integer values)
sub si, 2
SkipIf: cmp bx, si
jng RptLp
; Convert SI and BX back to I and J
lea ax, a
sub bx, ax
shr bx, 1
sub si, ax
shr si, 1
; Now for the recursive part:
mov ax, l
cmp ax, si
jnl NoRec1
push ax
push si
call sort
NoRec1: cmp bx, r
jnl NoRec2
push bx
push r
call sort
NoRec2: mov sp, bp
pop bp
ret 4
Sort endp
cseg ends
sseg segment stack 'stack'
word 256 dup (?)
sseg ends
end main
Other than some basic optimizations (like keeping several variables in registers),
this code is almost a literal translation of the Pascal code. Note that
the local variables i and j aren't necessary in
this assembly language code (we could use registers to hold their values).
Their use simply demonstrates the allocation of local variables on the stack.
sort in the assembly
language code above need not be a recursive call. By setting up a couple
of variables and registers, a simple jmp instruction can can
replace the pushes and the recursive call. This will improve the performance
of the quicksort routine (quite a bit, actually) and will reduce the amount
of memory the stack requires. A good book on algorithms, such as D.E. Knuth's
The Art of Computer Programming, Volume 3, would be an excellent source
of additional material on quicksort. Other texts on algorithm complexity,
recursion theory, and algorithms would be a good place to look for ideas
on efficiently implementing recursive algorithms.