The Art of
ASSEMBLY LANGUAGE PROGRAMMING

Chapter Nine (Part 4)

Table of Content

Chapter Nine (Part 6)

CHAPTER NINE:
ARITHMETIC AND LOGICAL OPERATIONS (Part 5)
9.3.6 - Extended Precision NEG Operations
9.3.7 - Extended Precision AND Operations
9.3.8 - Extended Precision OR Operations
9.3.9 - Extended Precision XOR Operations
9.3.10 - Extended Precision NOT Operations
9.3.11 - Extended Precision Shift Operations
9.3.12 - Extended Precision Rotate Operations
9.4 - Operating on Different Sized Operands

9.3.6 Extended Precision NEG Operations

Although there are several ways to negate an extended precision value the shortest way is to use a combination of neg and sbb instructions. This technique uses the fact that neg subtracts its operand from zero. In particular it sets the flags the same way the sub instruction would if you subtracted the destination value from zero. This code takes the following form:

                neg     dx
neg     ax
sbb     dx
0  

The sbb instruction decrements dx if there is a borrow out of the L.O. word of the negation operation (which always occurs unless ax is zero).

To extend this operation to additional bytes words or double words is easy; all you have to do is start with the H.O. memory location of the object you want to negate and work towards the L.O. byte. The following code computes a 128 bit negation on the 80386 processor:

Value           dword   0
0
0
0         ;128 bit integer.
.
.
.
neg     Value+12        ;Neg H.O. dword
neg     Value+8         ;Neg previous dword in memory.
sbb     Value+12
0     ;Adjust H.O. dword
neg     Value+4         ;Neg the second dword in object.
sbb     Value+8
0      ;Adjust 3rd dword in object.
sbb     Value+12
0     ;Carry any borrow through H.O. word.
neg     Value           ;Negate L.O. word.
sbb     Value+4
0      ;Adjust 2nd dword in object.
sbb     Value+8
0      ;Adjust 3rd dword in object.
sbb     Value+12
0     ;Carry any borrow through H.O. word.

Unfortunately this code tends to get really large and slow since you need to propogate the carry through all the H.O. words after each negate operation. A simpler way to negate larger values is to simply subract that value from zero:

Value           dword   0
0
0
0
0       ;160 bit integer.
.
.
.
mov     eax
0
sub     eax
Value
mov     Value
eax
mov     eax
0
sbb     eax
Value+4
mov     Value+8
ax
mov     eax
0
sbb     eax
Value+8
mov     Value+8
ax
mov     eax
0
sbb     eax
Value+12
mov     Value+12
ax
mov     eax
0
sbb     eax
Value+16
mov     Value+16
ax

9.3.7 Extended Precision AND Operations

Performing an n-word and operation is very easy - simply and the corresponding words between the two operands saving the result. For example to perform the and operation where all three operands are 32 bits long you could use the following code:

                mov     ax
word ptr source1
and     ax
word ptr source2
mov     word ptr dest
ax
mov     ax
word ptr source1+2
and     ax
word ptr source2+2
mov     word ptr dest+2
ax 

This technique easily extends to any number of words all you need to is logically and the corresponding bytes words or double words in the corresponding operands.

9.3.8 Extended Precision OR Operations

Multi-word logical or operations are performed in the same way as multi-word and operations. You simply or the corresponding words in the two operand together. For example to logically or two 48 bit values use the following code:

                mov     ax
word ptr operand1
or      ax
word ptr operand2
mov     word ptr operand3
ax
mov     ax
word ptr operand1+2
or      ax
word ptr operand2+2
mov     word ptr operand3+2
ax
mov     ax
word ptr operand1+4
or      ax
word ptr operand2+4
mov     word ptr operand3+4
ax

9.3.9 Extended Precision XOR Operations

Extended precision xor operations are performed in a manner identical to and/or - simply xor the corresponding words in the two operands to obtain the extended precision result. The following code sequence operates on two 64 bit operands computes their exclusive-or and stores the result into a 64 bit variable. This example uses the 32 bit registers available on 80386 and later processors.

                mov     eax
dword ptr operand1
xor     eax
dword ptr operand2
mov     dword ptr operand3
eax
mov     eax
dword ptr operand1+4
xor     eax
dword ptr operand2+4
mov     dword ptr operand3+4
eax

9.3.10 Extended Precision NOT Operations

The not instruction inverts all the bits in the specified operand. It does not affect any flags (therefore using a conditional jump after a not instruction has no meaning). An extended precision not is performed by simply executing the not instruction on all the affected operands. For example to perform a 32 bit not operation on the value in (dx:ax) all you need to do is execute the instructions:

        not     ax              or              not     dx
not     dx                              not     ax

Keep in mind that if you execute the not instruction twice you wind up with the original value. Also note that exclusive-oring a value with all ones (0FFh 0FFFFh or 0FF..FFh) performs the same operation as the not instruction.

9.3.11 Extended Precision Shift Operations

Extended precision shift operations require a shift and a rotate instruction. Consider what must happen to implement a 32 bit shl using 16 bit operations:

1) A zero must be shifted into bit zero.

2) Bits zero through 14 are shifted into the next higher bit.

3) Bit 15 is shifted into bit 16.

4) Bits 16 through 30 must be shifted into the next higher bit.

5) Bit 31 is shifted into the carry flag.

The two instructions you can use to implement this 32 bit shift are shl and rcl. For example to shift the 32 bit quantity in (dx:ax) one position to the left you'd use the instructions:

                shl     ax
1
rcl     dx
1

Note that you can only shift an extended precision value one bit at a time. You cannot shift an extended precision operand several bits using the cl register or an immediate value greater than one as the count using this technique.

To understand how this instruction sequence works consider the operation of these instructions on an individual basis. The shl instruction shifts a zero into bit zero of the 32 bit operand and shifts bit 15 into the carry flag. The rcl instruction then shifts the carry flag into bit 16 and then shifts bit 31 into the carry flag. The result is exactly what we want.

To perform a shift left on an operand larger than 32 bits you simply add additional rcl instructions. An extended precision shift left operation always starts with the least significant word and each succeeding rcl instruction operates on the next most significant word. For example to perform a 48 bit shift left operation on a memory location you could use the following instructions:

                shl     word ptr Operand
1
rcl     word ptr Operand+2
1
rcl     word ptr Operand+4
1

If you need to shift your data by two or more bits you can either repeat the above sequence the desired number of times (for a constant number of shifts) or you can place the instructions in a loop to repeat them some number of times. For example the following code shifts the 48 bit value Operand to the left the number of bits specified in cx:

ShiftLoop:      shl     word ptr Operand
1
rcl     word ptr Operand+2
1
rcl     word ptr Operand+4
1
loop    ShiftLoop

You implement shr and sar in a similar way except you must start at the H.O. word of the operand and work your way down to the L.O. word:

DblSAR:         sar word ptr Operand+4
1
rcr word ptr Operand+2
1
rcr word ptr Operand
1

DblSHR:         shr word ptr Operand+4
1
rcr word ptr Operand+2
1
rcr word ptr Operand
1

There is one major difference between the extended precision shifts described here and their 8/16 bit counterparts - the extended precision shifts set the flags differently than the single precision operations. For example the zero flag is set if the last rotate instruction produced a zero result not if the entire shift operation produced a zero result. For the shift right operations the overflow and sign flags aren't set properly (they are set properly for the left shift operation). Additional testing will be required if you need to test one of these flags after an extended precision shift operation. Fortunately the carry flag is the flag most often tested after a shift operation and the extended precision shift instructions properly set this flag.

The shld and shrd instructions let you efficiently implement multiprecision shifts of several bits on 80386 and later processors. Consider the following code sequence:

ShiftMe         dword   1234h
5678h
9012h
.
.
.
mov     eax
ShiftMe+4
shld    ShiftMe+8
eax
6
mov     eax
ShiftMe
shld    ShiftMe+4
eax
6
shl     ShiftMe
6

Recall that the shld instruction shifts bits from its second operand into its first operand. Therefore the first shld instruction above shifts the bits from ShiftMe+4 into ShiftMe+8 without affecting the value in ShiftMe+4. The second shld instruction shifts the bits from ShiftMe into ShiftMe+4. Finally the shl instruction shifts the L.O. double word the appropriate amount. There are two important things to note about this code. First unlike the other extended precision shift left operations this sequence works from the H.O. double word down to the L.O. double word. Second the carry flag does not contain the carry out of the H.O. shift operation. If you need to preserve the carry flag at that point you will need to push the flags after the first shld instruction and pop the flags after the shl instruction.

You can do an extended precision shift right operation using the shrd instruction. It works almost the same way as the code sequence above except you work from the L.O. double word to the H.O. double word. The solution is left as an exercise at the end of this chapter.

9.3.12 Extended Precision Rotate Operations

The rcl and rcr operations extend in a manner almost identical to that for shl and shr . For example to perform 48 bit rcl and rcr operations use the following instructions:

                rcl     word ptr operand
1
rcl     word ptr operand+2
1
rcl     word ptr operand+4
1

rcr     word ptr operand+4
1
rcr     word ptr operand+2
1
rcr     word ptr operand
1

The only difference between this code and the code for the extended precision shift operations is that the first instruction is a rcl or rcr rather than a shl or shr instruction.

Performing an extended precision rol or ror instruction isn't quite as simple an operation. The 8086 extended precision versions of these instructions appear in the exercises. On the 80386 and later processors you can use the bt shld and shrd instructions to easily implement an extended precision rol or ror instruction. The following code shows how to use the shld instruction to do an extended precision rol:

; Compute ROL EDX:EAX
4

mov     ebx
edx
shld    edx
eax
4
shld    eax
ebx
4
bt      eax
0          ;Set carry flag
if desired.

An extended precision ror instruction is similar; just keep in mind that you work on the L.O. end of the object first and the H.O. end last.

9.4 Operating on Different Sized Operands

Occasionally you may need to compute some value on a pair of operands that are not the same size. For example you may need to add a word and a double word together or subtract a byte value from a word value. The solution is simple: just extend the smaller operand to the size of the larger operand and then do the operation on two similarly sized operands. For signed operands you would sign extend the smaller operand to the same size as the larger operand; for unsigned values you zero extend the smaller operand. This works for any operation although the following examples demonstrate this for the addition operation.

To extend the smaller operand to the size of the larger operand use a sign extension or zero extension operation (depending upon whether you're adding signed or unsigned values). Once you've extended the smaller value to the size of the larger the addition can proceed. Consider the following code that adds a byte value to a word value:

var1            byte    ?
var2            word    ?

Unsigned addition:                      Signed addition:

mov     al
var1                        mov     al
var1
mov     ah
0                           cbw
add     ax
var2                        add     ax
var2

In both cases the byte variable was loaded into the al register extended to 16 bits and then added to the word operand. This code works out really well if you can choose the order of the operations (e.g. adding the eight bit value to the sixteen bit value). Sometimes you cannot specify the order of the operations. Perhaps the sixteen bit value is already in the ax register and you want to add an eight bit value to it. For unsigned addition you could use the following code:

                mov     ax
var2                ;Load 16 bit value into AX
.                      ;Do some other operations leaving
.                      ; a 16 bit quantity in AX.
add     al
var1        ;Add in the 8 bit value.
adc     ah
0           ;Add carry into the H.O. word.

The first add instruction in this example adds the byte at var1 to the L.O. byte of the value in the accumulator. The adc instruction above adds the carry out of the L.O. byte into the H.O. byte of the accumulator. Care must be taken to ensure that this adc instruction is present. If you leave it out you may not get the correct result.

Adding an eight bit signed operand to a sixteen bit signed value is a little more difficult. Unfortunately you cannot add an immediate value (as above) to the H.O. word of ax. This is because the H.O. extension byte can be either 00h or 0FFh. If a register is available the best thing to do is the following:

                mov     bx
ax          ;BX is the available register.
mov     al
var1
cbw
add     ax
bx

If an extra register is not available you might try the following code:

                add     al
var1
cmp     var1
0
jge     add0
adc     ah
0FFh
jmp     addedFF
add0:           adc     ah
0
addedFF:

Of course if another register isn't available you could always push one onto the stack and save it while you're performing the operation e.g.

                push    bx
mov     bx
ax
mov     al
var1
cbw
add     ax
bx
pop     bx

Another alternative is to store the 16 bit value in the accumulator into a memory location and then proceed as before:

                mov     temp
ax
mov     al
var1
cbw
add     ax
temp

All the examples above added a byte value to a word value. By zero or sign extending the smaller operand to the size of the larger operand you can easily add any two different sized variables together. Consider the following code that adds a signed byte operand to a signed double word:

var1            byte    ?
var2            dword   ?

mov     al
var1
cbw
cwd                             ;Extend to 32 bits in DX
add     ax
word ptr var2
adc     dx
word ptr var2+2

Of course if you have an 80386 or later processor you could use the following code:

                movsx   eax
var1
add     eax
var2

An example more applicable to the 80386 is adding an eight bit value to a quadword (64 bit) value consider the following code:

BVal            byte    -1
QVal            qword   1

movsx   eax
BVal
cdq
add     eax
dword ptr QVal
adc     edx
dword ptr QVal+4

For additional examples see the exercises at the end of this chapter.

Chapter Nine (Part 4)

Table of Content

Chapter Nine (Part 6)

Chapter Nine: Arithmetic And Logical Operations (Part 5)
27 SEP 1996