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Art of Assembly: Chaper Eleven


11.5.9 - Passing Parameters on the Stack

11.5.9 Passing Parameters on the Stack


Most HLLs use the stack to pass parameters because this method is fairly efficient. To pass parameters on the stack, push them immediately before calling the subroutine. The subroutine then reads this data from the stack memory and operates on it appropriately. Consider the following Pascal procedure call:




		CallProc(i,j,k+4);

Most Pascal compilers push their parameters onto the stack in the order that they appear in the parameter list. Therefore, the 80x86 code typically emitted for this subroutine call (assuming you're passing the parameters by value) is





                push    i
                push    j
                mov     ax, k
                add     ax, 4
                push    ax
                call    CallProc

Upon entry into CallProc, the 80x86's stack looks like that shown below (for a near or a far procedure).

You could gain access to the parameters passed on the stack by removing the data from the stack (Assuming a near procedure call):





CallProc        proc    near
                pop     RtnAdrs
                pop     kParm
                pop     jParm
                pop     iParm
                push    RtnAdrs
                 .
                 .
                 .
                ret
CallProc        endp

There is, however, a better way. The 80x86's architecture allows you to use the bp (base pointer) register to access parameters passed on the stack. This is one of the reasons the disp[bp], [bp][di], [bp][si], disp[bp][si], and disp[bp][di] addressing modes use the stack segment rather than the data segment. The following code segment gives the standard procedure entry and exit code:





StdProc         proc    near
                push    bp
                mov     bp, sp
                 .
                 .
                 .
                pop     bp
                ret     ParmSize
StdProc         endp

ParmSize is the number of bytes of parameters pushed onto the stack before calling the procedure. In the CallProc procedure there were six bytes of parameters pushed onto the stack so ParmSize would be six.

Take a look at the stack immediately after the execution of mov bp, sp in StdProc. Assuming you've pushed three parameter words onto the stack, it should look something like shown below:

Now the parameters can be fetched by indexing off the bp register:





                mov     ax, 8[bp]        ;Accesses the first parameter
                mov     ax, 6[bp]        ;Accesses the second parameter
                mov     ax, 4[bp]        ;Accesses the third parameter

When returning to the calling code, the procedure must remove these parameters from the stack. To accomplish this, pop the old bp value off the stack and execute a ret 6 instruction. This pops the return address off the stack and adds six to the stack pointer, effectively removing the parameters from the stack.

The displacements given above are for near procedures only. When calling a far procedure,




The stack contents when calling a far procedure are shown below:

This collection of parameters, return address, registers saved on the stack, and other items, is a stack frame or activation record.

When saving other registers onto the stack, always make sure that you save and set up bp before pushing the other registers. If you push the other registers before setting up bp, the offsets into the stack frame will change. For example, the following code disturbs the ordering presented above:





FunnyProc       proc    near
                push    ax
                push    bx
                push    bp
                mov     bp, sp
                 .
                 .
                 .
                pop     bp
                pop     bx
                pop     ax
                ret
FunnyProc       endp

Since this code pushes ax and bx before pushing bp and copying sp to bp, ax and bx appear in the activation record before the return address (that would normally start at location [bp+2]). As a result, the value of bx appears at location [bp+2] and the value of ax appears at location [bp+4]. This pushes the return address and other parameters farther up the stack as shown below:

Although this is a near procedure, the parameters don't begin until offset eight in the activation record. Had you pushed the ax and bx registers after setting up bp, the offset to the parameters would have been four:





FunnyProc       proc    near
                push    bp
                mov     bp, sp
                push    ax
                push    bx
                 .
                 .
                 .
                pop     bx
                pop     ax
                pop     bp
                ret
FunnyProc       endp

Therefore, the push bp and mov bp, sp instructions should be the first two instructions any subroutine executes when it has parameters on the stack.

Accessing the parameters using expressions like [bp+6] can make your programs very hard to read and maintain. If you would like to use meaningful names, there are several ways to do so. One way to reference parameters by name is to use equates. Consider the following Pascal procedure and its equivalent 80x86 assembly language code:





        procedure xyz(var i:integer; j,k:integer);
        begin
                i := j+k;
        end;

Calling sequence:





                xyz(a,3,4);

Assembly language code:





xyz_i           equ     8[bp]           ;Use equates so we can reference
xyz_j           equ     6[bp]           ; symbolic names in the body of
xyz_k           equ     4[bp]           ; the procedure.
xyz             proc    near
                push    bp
                mov     bp, sp
                push    es
                push    ax
                push    bx
                les     bx, xyz_i       ;Get address of I into ES:BX
                mov     ax, xyz_j       ;Get J parameter
                add     ax, xyz_k       ;Add to K parameter
                mov     es:[bx], ax     ;Store result into I parameter
                pop     bx
                pop     ax
                pop     es
                pop     bp
                ret     8
xyz             endp

Calling sequence:





                mov     ax, seg a       ;This parameter is passed by
                push    ax              ; reference, so pass its
                mov     ax, offset a    ; address on the stack.
                push    ax
                mov     ax, 3           ;This is the second parameter
                push    ax
                mov     ax, 4           ;This is the third parameter.
                push    ax
                call    xyz

On an 80186 or later processor you could use the following code in place of the above:





                push    seg a           ;Pass address of "a" on the
                push    offset a        ; stack.
                push    3               ;Pass second parm by val.
                push    4               ;Pass third parm by val.
                call    xyz

Upon entry into the xyz procedure, before the execution of the les instruction, the stack looks like shown below:

Since you're passing I by reference, you must push its address onto the stack. This code passes reference parameters using 32 bit segmented addresses. Note that this code uses ret 8. Although there are three parameters on the stack, the reference parameter I consumes four bytes since it is a far address. Therefore there are eight bytes of parameters on the stack necessitating the ret 8 instruction.

Were you to pass I by reference using a near pointer rather than a far pointer, the code would look like the following:





xyz_i           equ     8[bp]           ;Use equates so we can reference
xyz_j           equ     6[bp]           ; symbolic names in the body of
xyz_k           equ     4[bp]           ; the procedure.
xyz             proc    near
                push    bp
                mov     bp, sp
                push    ax
                push    bx
                mov     bx, xyz_i        ;Get address of I into BX
                mov     ax, xyz_j        ;Get J parameter
                add     ax, xyz_k        ;Add to K parameter
                mov     [bx], ax        ;Store result into I parameter
                pop     bx
                pop     ax
                pop     bp
                ret     6
xyz             endp

Note that since I's address on the stack is only two bytes (rather than four), this routine only pops six bytes when it returns.

Calling sequence:





                mov     ax, offset a    ;Pass near address of a.
                push    ax
                mov     ax, 3           ;This is the second parameter
                push    ax
                mov     ax, 4           ;This is the third parameter.
                push    ax
                call    xyz

On an 80286 or later processor you could use the following code in place of the above:





                push    offset a        ;Pass near address of a.
                push    3               ;Pass second parm by val.
                push    4               ;Pass third parm by val.
                call    xyz

The stack frame for the above code appears below:

When passing a parameter by value-returned or result, you pass an address to the procedure, exactly like passing the parameter by reference. The only difference is that you use a local copy of the variable within the procedure rather than accessing the variable indirectly through the pointer. The following implementations for xyz show how to pass I by value-returned and by result:





; xyz version using Pass by Value-Returned for xyz_i

xyz_i           equ     8[bp]           ;Use equates so we can reference
xyz_j           equ     6[bp]           ; symbolic names in the body of
xyz_k           equ     4[bp]           ; the procedure.

xyz             proc    near
                push    bp
                mov     bp, sp
                push    ax
                push    bx
                push    cx              ;Keep local copy here.

                mov     bx, xyz_i        ;Get address of I into BX
                mov     cx, [bx]        ;Get local copy of I parameter.

                mov     ax, xyz_j        ;Get J parameter
                add     ax, xyz_k        ;Add to K parameter
                mov     cx, ax          ;Store result into local copy

                mov     bx, xyz_i       ;Get ptr to I, again
                mov     [bx], cx        ;Store result away.

                pop     cx
                pop     bx
                pop     ax
                pop     bp
                ret     6
xyz             endp

There are a couple of unnecessary mov instructions in this code. They are present only to precisely implement pass by value-returned parameters. It is easy to improve this code using pass by result parameters. The modified code is





; xyz version using Pass by Result for xyz_i

xyz_i           equ     8[bp]           ;Use equates so we can reference
xyz_j           equ     6[bp]           ; symbolic names in the body of
xyz_k           equ     4[bp]           ; the procedure.

xyz             proc    near
                push    bp
                mov     bp, sp
                push    ax
                push    bx
                push    cx              ;Keep local copy here.

                mov     ax, xyz_j        ;Get J parameter
                add     ax, xyz_k        ;Add to K parameter
                mov     cx, ax          ;Store result into local copy

                mov     bx, xyz_i       ;Get ptr to I, again
                mov     [bx], cx        ;Store result away.

                pop     cx
                pop     bx
                pop     ax
                pop     bp
                ret     6
xyz             endp

As with passing value-returned and result parameters in registers, you can improve the performance of this code using a modified form of pass by value. Consider the following implementation of xyz:





; xyz version using modified pass by value-result for xyz_i

xyz_i           equ     8[bp]           ;Use equates so we can reference
xyz_j           equ     6[bp]           ; symbolic names in the body of
xyz_k           equ     4[bp]           ; the procedure.

xyz             proc    near
                push    bp
                mov     bp, sp
                push    ax

                mov     ax, xyz_j       ;Get J parameter
                add     ax, xyz_k       ;Add to K parameter
                mov     xyz_i, ax       ;Store result into local copy

                pop     ax
                pop     bp
                ret     4               ;Note that we do not pop I parm.
xyz             endp

The calling sequence for this code is
                push    a               ;Pass a's value to xyz.
                push    3               ;Pass second parameter by val.
                push    4               ;Pass third parameter by val.
                call    xyz
                pop     a

Note that a pass by result version wouldn't be practical since you have to push something on the stack to make room for the local copy of I inside xyz. You may as well push the value of a on entry even though the xyz procedure ignores it. This procedure pops only four bytes off the stack on exit. This leaves the value of the I parameter on the stack so that the calling code can store it away to the proper destination.

To pass a parameter by name on the stack, you simply push the address of the thunk. Consider the following pseudo-Pascal code:





procedure swap(name Item1, Item2:integer);
var temp:integer;
begin

        temp := Item1;
        Item1 := Item2;
        Item2 := Temp;

end;

If swap is a near procedure, the 80x86 code for this procedure could look like the following (note that this code has been slightly optimized and does not following the exact sequence given above):





; swap-         swaps two parameters passed by name on the stack.
;               Item1 is passed at address [bp+6], Item2 is passed
;               at address [bp+4]

wp              textequ <word ptr>
swap_Item1      equ     [bp+6]
swap_Item2      equ     [bp+4]

swap            proc    near
                push    bp
                mov     bp, sp
                push    ax                      ;Preserve temp value.
                push    bx                      ;Preserve bx.
                call    wp swap_Item1           ;Get adrs of Item1.
                mov     ax, [bx]                ;Save in temp (AX).
                call    wp swap_Item2           ;Get adrs of Item2.
                xchg    ax, [bx]                ;Swap temp <-> Item2.
                call    wp swap_Item1           ;Get adrs of Item1.
                mov     [bx], ax                ;Save temp in Item1.
                pop     bx                      ;Restore bx.
                pop     ax                      ;Restore ax.
                ret     4                       ;Return and pop Item1/2.
swap            endp

Some sample calls to swap follow:





; swap(A[i], i) -- 8086 version.

                lea     ax, thunk1
                push    ax
                lea     ax, thunk2
                push    ax
                call    swap

; swap(A[i],i) -- 80186 & later version.

                push    offset thunk1
                push    offset thunk2
                call    swap

                 .
                 .
                 .

; Note: this code assumes A is an array of two byte integers.

thunk1          proc    near
                mov     bx, i
                shl     bx, 1
                lea     bx, A[bx]
                ret
thunk1          endp

thunk2          proc    near
                lea     bx, i
                ret
thunk2          endp

The code above assumes that the thunks are near procs that reside in the same segment as the swap routine. If the thunks are far procedures the caller must pass far addresses on the stack and the swap routine must manipulate far addresses. The following implementation of swap, thunk1, and thunk2 demonstrate this.





; swap-         swaps two parameters passed by name on the stack.
;               Item1 is passed at address [bp+10], Item2 is passed
;               at address [bp+6]

swap_Item1      equ     [bp+10]
swap_Item2      equ     [bp+6]
dp              textequ <dword ptr>

swap            proc    far
                push    bp
                mov     bp, sp
                push    ax              ;Preserve temp value.
                push    bx              ;Preserve bx.
                push    es              ;Preserve es.
                call    dp swap_Item1   ;Get adrs of Item1.
                mov     ax, es:[bx]     ;Save in temp (AX).
                call    dp swap_Item2   ;Get adrs of Item2.
                xchg    ax, es:[bx]     ;Swap temp <-> Item2.
                call    dp swap_Item1   ;Get adrs of Item1.
                mov     es:[bx], ax     ;Save temp in Item1.
                pop     es              ;Restore es.
                pop     bx              ;Restore bx.
                pop     ax              ;Restore ax.
                ret     8               ;Return and pop Item1, Item2.
swap            endp

Some sample calls to swap follow:





; swap(A[i], i) -- 8086 version.

                mov     ax, seg thunk1
                push    ax
                lea     ax, thunk1
                push    ax
                mov     ax, seg thunk2
                push    ax
                lea     ax, thunk2
                push    ax
                call    swap

; swap(A[i],i) -- 80186 & later version.

                push    seg thunk1
                push    offset thunk1
                push    seg thunk2
                push    offset thunk2
                call    swap

                 .
                 .
                 .

; Note:         this code assumes A is an array of two byte integers.
;               Also note that we do not know which segment(s) contain
;               A and I.

thunk1          proc    far
                mov     bx, seg A       ;Need to return seg A in ES.
                push    bx              ;Save for later.
                mov     bx, seg i       ;Need segment of I in order
                mov     es, bx          ; to access it.
                mov     bx, es:i        ;Get I's value.
                shl     bx, 1
                lea     bx, A[bx]
                pop     es              ;Return segment of A[I] in es.
                ret
thunk1          endp

thunk2          proc    near
                mov     bx, seg i       ;Need to return I's seg in es.
                mov     es, bx
                lea     bx, i
                ret
thunk2          endp

Passing parameters by lazy evaluation is left for the programming projects.

Additional information on activation records and stack frames appears later in this chapter in the section on local variables.

11.5.9 - Passing Parameters on the Stack


Art of Assembly: Chaper Eleven - 27 SEP 1996

[Chapter Eleven][Previous] [Next] [Art of Assembly][Randall Hyde]



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