There is a lot more to assembly language than knowing the operations of a handful of machine instructions. You've got to know how to use them and what they can do. Many instructions are useful for operations that have little to do with their mathematical or obvious functions. This chapter discusses how to convert expressions from a high level language into assembly language. It also discusses advanced arithmetic and logical operations including multiprecision operations and tricks you can play with various instructions.

This chapter discusses six main subjects: converting HLL arithmetic expressions into assembly language logical expressions extended precision arithmetic and logical operations operating on different sized operands machine and arithmetic idioms and masking operations. Like the preceding chapters this chapter contains considerable material that you may need to learn immediately if you're a beginning assembly language programmer. The sections below that have a "*" prefix are essential. Those sections with a "o" discuss advanced topics that you may want to put off for a while.

* Arithmetic expressions

* Simple assignments

* Simple expressions

* Complex expressions

* Commutative operators

* Logical expressions

* Multiprecision operations

* Multiprecision addition operations

* Multiprecision subtraction operations

* Extended precision comparisons

    o Extended precision multiplication

    o Extended precision division

    o Extended precision negation

* Extended precision AND OR XOR and NOT

    o Extended precision shift and rotate operations

    o Operating on different sized operands

* Multiplying without MUL and IMUL

    o Division without DIV and IDIV

    o Using AND to compute remainders

    o Modulo-n Counters with AND

    o Testing for 0FFFFF...FFFh

* Test operations

    o Testing signs with the XOR instructions

    o Masking operations

    o Masking with the AND instructions

    o Masking with the OR instruction

    o Packing and unpacking data types

    o Table lookups

None of this material is particularly difficult to understand. However there are a lot of new topics here and taking them a few at a time will certain help you absorb the material better. Those topics with the "*" prefix are ones you will frequently use; hence it is a good idea to study these first.

Probably the biggest shock to beginners facing assembly language for the very first time is the lack of familiar arithmetic expressions. Arithmetic expressions in most high level languages look similar to their algebraic equivalents:

		X:=Y*Z;

In assembly language you'll need several statements to accomplish this same task e.g.

                mov     ax
y
imul    z
mov     x
ax

Obviously the HLL version is much easier to type read and understand. This point more than any other is responsible for scaring people away from assembly language.

Although there is a lot of typing involved converting an arithmetic expression into assembly language isn't difficult at all. By attacking the problem in steps the same way you would solve the problem by hand you can easily break down any arithmetic expression into an equivalent sequence of assembly language statements. By learning how to convert such expressions to assembly language in three steps you'll discover there is little difficulty to this task.

9.1.1 Simple Assignments

The easiest expressions to convert to assembly language are the simple assignments. Simple assignments copy a single value into a variable and take one of two forms:

		variable := constant 

or

		variable := variable

If variable appears in the current data segment (e.g. DSEG) converting the first form to assembly language is easy just use the assembly language statement:

                  mov    variable
constant

This move immediate instruction copies the constant into the variable.

The second assignment above is somewhat complicated since the 80x86 doesn't provide a memory-to-memory mov instruction. Therefore to copy one memory variable into another you must move the data through a register. If you'll look at the encoding for the mov instruction in the appendix you'll notice that the mov ax memory and mov memory ax instructions are shorter than moves involving other registers. Therefore if the ax register is available you should use it for this operation. For example

		var1 := var2; 

becomes

                mov     ax
var2
mov     var1
ax

Of course if you're using the ax register for something else one of the other registers will suffice. Regardless you must use a register to transfer one memory location to another.

This discussion of course assumes that both variables are in memory. If possible you should try to use a register to hold the value of a variable.

9.1.2 Simple Expressions

The next level of complexity up from a simple assignment is a simple expression. A simple expression takes the form:

		var := term1 op term2;

Var is a variable term1 and term2 are variables or constants and op is some arithmetic operator (addition subtraction multiplication etc.).

As simple as this expression appears most expressions take this form. It should come as no surprise then that the 80x86 architecture was optimized for just this type of expression.

A typical conversion for this type of expression takes the following form:

                mov     ax
term1
op      ax
term2
mov     var
ax

Op is the mnemonic that corresponds to the specified operation (e.g. "+" = add "-" = sub etc.).

There are a few inconsistencies you need to be aware of. First the 80x86's {i}mul instructions do not allow immediate operands on processors earlier than the 80286. Further none of the processors allow immediate operands with {i}div. Therefore if the operation is multiplication or division and one of the terms is a constant value you may need to load this constant into a register or memory location and then multiply or divide ax by that value. Of course when dealing with the multiply and divide instructions on the 8086/8088 you must use the ax and dx registers. You cannot use arbitrary registers as you can with other operations. Also don't forget the sign extension instructions if you're performing a division operation and you're dividing one 16/32 bit number by another. Finally don't forget that some instructions may cause overflow. You may want to check for an overflow (or underflow) condition after an arithmetic operation.

Examples of common simple expressions:

X := Y + Z;

mov     ax
y
add     ax
z
mov     x
ax

X := Y - Z;

mov     ax
y
sub     ax
z
mov     x
ax

X := Y * Z; {unsigned}

mov     ax
y
mul     z               ;Use IMUL for signed arithmetic.
mov     x
ax           ;Don't forget this wipes out DX.

X := Y div Z; {unsigned div}

mov     ax
y
mov     dx
0           ;Zero extend AX into DX
div     z
mov     x
ax

X := Y div Z; {signed div}

mov     ax
y
cwd                     ;Sign extend AX into DX
idiv    z
mov     x
ax

X := Y mod Z; {unsigned remainder}

mov     ax
y
mov     dx
0           ;Zero extend AX into DX
div     z
mov     x
dx           ;Remainder is in DX

X := Y mod Z; {signed remainder}

mov     ax
y
cwd                     ;Sign extend AX into DX
idiv    z
mov     x
dx           ;Remainder is in DX

Since it is possible for an arithmetic error to occur you should generally test the result of each expression for an error before or after completing the operation. For example unsigned addition subtraction and multiplication set the carry flag if an overflow occurs. You can use the jc or jnc instructions immediately after the corresponding instruction sequence to test for overflow. Likewise you can use the jo or jno instructions after these sequences to test for signed arithmetic overflow. The next two examples demonstrate how to do this for the add instruction:

X := Y + Z; {unsigned}

mov     ax
y
add     ax
z
mov     x
ax
jc      uOverflow

X := Y + Z; {signed}

mov     ax
y
add     ax
z
mov     x
ax
jo      sOverflow

Certain unary operations also qualify as simple expressions. A good example of a unary operation is negation. In a high level language negation takes one of two possible forms:

var := -var			or		var1 := -var2

Note that var := -constant is really a simple assignment not a simple expression. You can specify a negative constant as an operand to the mov instruction:

                 mov      var
-14

To handle the first form of the negation operation above use the single assembly language statement:

                 neg      var

If two different variables are involved then use the following:

                mov     ax
var2
neg     ax
mov     var1
ax

Overflow only occurs if you attempt to negate the most negative value (-128 for eight bit values -32768 for sixteen bit values etc.). In this instance the 80x86 sets the overflow flag so you can test for arithmetic overflow using the jo or jno instructions. In all other cases the80x86 clears the overflow flag. The carry flag has no meaning after executing the neg instruction since neg (obviously) does not apply to unsigned operands.

9.1.3 Complex Expressions

A complex expression is any arithmetic expression involving more than two terms and one operator. Such expressions are commonly found in programs written in a high level language. Complex expressions may include parentheses to override operator precedence function calls array accesses etc. While the conversion of some complex expressions to assembly language is fairly straight-forward others require some effort. This section outlines the rules you use to convert such expressions.

A complex function that is easy to convert to assembly language is one that involves three terms and two operators for example:

		W := W - Y - Z;

Clearly the straight-forward assembly language conversion of this statement will require two sub instructions. However even with an expression as simple as this one the conversion is not trivial. There are actually two ways to convert this from the statement above into assembly language:

                mov     ax
w
sub     ax
y
sub     ax
z
mov     w
ax

and

                mov     ax
y
sub     ax
z
sub     w
ax

The second conversion since it is shorter looks better. However it produces an incorrect result (assuming Pascal-like semantics for the original statement). Associativity is the problem. The second sequence above computes W := W - (Y - Z) which is not the same as W := (W - Y) - Z. How we place the parentheses around the subexpressions can affect the result. Note that if you are interested in a shorter form you can use the following sequence:

                mov     ax
y
add     ax
z
sub     w
ax

This computes W:=W-(Y+Z). This is equivalent to W := (W - Y) - Z.

Precedence is another issue. Consider the Pascal expression:

		X := W * Y + Z;

Once again there are two ways we can evaluate this expression:

		X := (W * Y) + Z;
or
		X := W * (Y + Z);

By now you're probably thinking that this text is crazy. Everyone knows the correct way to evaluate these expressions is the second form provided in these two examples. However you're wrong to think that way. The APL programming language for example evaluates expressions solely from right to left and does not give one operator precedence over another.

Most high level languages use a fixed set of precedence rules to describe the order of evaluation in an expression involving two or more different operators. Most programming languages for example compute multiplication and division before addition and subtraction. Those that support exponentiation (e.g. FORTRAN and BASIC) usually compute that before multiplication and division. These rules are intuitive since almost everyone learns them before high school. Consider the expression:

		X op1 Y op2 Z

If op1 takes precedence over op2 then this evaluates to (X op1 Y) op2 Z otherwise if op2 takes precedence over op1 then this evaluates to X op1 (Y op2 Z ). Depending upon the operators and operands involved these two computations could produce different results.

When converting an expression of this form into assembly language you must be sure to compute the subexpression with the highest precedence first. The following example demonstrates this technique:

; W := X + Y * Z;

mov     bx
x
mov     ax
y           ;Must compute Y * Z first since
mul     z               ; "*" has the highest precedence.
add     bx
ax          ;Now add product with X's value.
mov     w
bx           ;Save away result.

Since addition is a commutative operation we could optimize the above code to produce:

; W := X + Y * Z;

mov     ax
y           ;Must compute Y * Z first since
mul     z               ; "*" has the highest precedence.
add     ax
x           ;Now add product with X's value.
mov     w
ax           ;Save away result.

If two operators appearing within an expression have the same precedence then you determine the order of evaluation using associativity rules. Most operators are left associative meaning that they evaluate from left to right. Addition subtraction multiplication and division are all left associative. A right associative operator evaluates from right to left. The exponentiation operator in FORTRAN and BASIC is a good example of a right associative operator:

		2^2^3 is equal to 2^(2^3) not (2^2)^3

The precedence and associativity rules determine the order of evaluation. Indirectly these rules tell you where to place parentheses in an expression to determine the order of evaluation. Of course you can always use parentheses to override the default precedence and associativity. However the ultimate point is that your assembly code must complete certain operations before others to correctly compute the value of a given expression. The following examples demonstrate this principle:

; W := X - Y - Z

mov     ax
x           ;All the same operator
so we need
sub     ax
y           ; to evaluate from left to right
sub     ax
z           ; because they all have the same
mov     w
ax           ; precedence.

; W := X + Y * Z

mov     ax
y           ;Must compute Y * Z first since
imul    z               ; multiplication has a higher
add     ax
x           ; precedence than addition.
mov     w
ax

; W := X / Y - Z

mov     ax
x           ;Here we need to compute division
cwd                     ; first since it has the highest
idiv    y               ; precedence.
sub     ax
z
mov     w
ax

; W := X * Y * Z

mov     ax
y           ;Addition and multiplication are
imul    z               ; commutative
therefore the order
imul    x               ; of evaluation does not matter
mov     w
ax

There is one exception to the associativity rule. If an expression involves multiplication and division it is always better to perform the multiplication first. For example given an expression of the form:

		W := X/Y * Z

It is better to compute X*Z and then divide the result by Y rather than divide X by Y and multiply the quotient by Z. There are two reasons this approach is better. First remember that the imul instruction always produces a 32 bit result (assuming 16 bit operands). By doing the multiplication first you automatically sign extend the product into the dx register so you do not have to sign extend ax prior to the division. This saves the execution of the cwd instruction. A second reason for doing the multiplication first is to increase the accuracy of the computation. Remember (integer) division often produces an inexact result. For example if you compute 5/2 you will get the value two not 2.5. Computing (5/2)*3 produces six. However if you compute (5*3)/2 you get the value seven which is a little closer to the real quotient (7.5). Therefore if you encounter an expression of the form:

		W := X/Y*Z;

You can usually convert this to assembly code:

                mov     ax
x
imul    z
idiv    z
mov     w
ax

Of course if the algorithm you're encoding depends on the truncation effect of the division operation you cannot use this trick to improve the algorithm. Moral of the story: always make sure you fully understand any expression you are converting to assembly language. Obviously if the semantics dictate that you must perform the division first do so.

Consider the following Pascal statement:

		W := X - Y * Z;

This is similar to a previous example except it uses subtraction rather than addition. Since subtraction is not commutative you cannot compute Y * Z and then subtract X from this result. This tends to complicate the conversion a tiny amount. Rather than a straight forward multiply and addition sequence you'll have to load X into a register multiply Y and Z leaving their product in a different register and then subtract this product from X e.g.

                mov     bx
x
mov     ax
y
imul    z
sub     bx
ax
mov     w
bx

This is a trivial example that demonstrates the need for temporary variables in an expression. The code uses the bx register to temporarily hold a copy of X until it computes the product of Y and Z. As your expression increase in complexity the need for temporaries grows. Consider the following Pascal statement:

		W := (A + B) * (Y + Z);

Following the normal rules of algebraic evaluation you compute the subexpressions inside the parentheses (i.e. the two subexpressions with the highest precedence) first and set their values aside. When you computed the values for both subexpressions you can compute their sum. One way to deal with complex expressions like this one is to reduce it to a sequence of simple expressions whose results wind up in temporary variables. For example we can convert the single expression above into the following sequence:

		Temp1 := A + B;
Temp2 := Y + Z;
W := Temp1 * Temp2;

Since converting simple expressions to assembly language is quite easy it's now a snap to compute the former complex expression in assembly. The code is

                mov     ax
a
add     ax
b
mov     Temp1
ax
mov     ax
y
add     ax
z
mov     temp2
ax
mov     ax
temp1

imul    temp2
mov     w
ax

Of course this code is grossly inefficient and it requires that you declare a couple of temporary variables in your data segment. However it is very easy to optimize this code by keeping temporary variables as much as possible in 80x86 registers. By using 80x86 registers to hold the temporary results this code becomes:

                mov     ax
a
add     ax
b
mov     bx
y
add     bx
z
imul    bx
mov     w
ax

Yet another example:

                X := (Y+Z) * (A-B) / 10;

This can be converted to a set of four simple expressions:

                Temp1 := (Y+Z)
Temp2 := (A-B)
Temp1 := Temp1 * Temp2
X := Temp1 / 10

You can convert these four simple expressions into the assembly language statements:

                mov     ax
y           ;Compute AX := Y+Z
add     ax
z
mov     bx
a           ;Compute BX := A-B
sub     bx
b
mul     bx              ;Compute AX := AX * BX
this also sign
mov     bx
10          ; extends AX into DX for idiv.
idiv    bx              ;Compute AX := AX / 10
mov     x
ax           ;Store result into X

The most important thing to keep in mind is that temporary values if possible should be kept in registers. Remember accessing an 80x86 register is much more efficient than accessing a memory location. Use memory locations to hold temporaries only if you've run out of registers to use.

Ultimately converting a complex expression to assembly language is little different than solving the expression by hand. Instead of actually computing the result at each stage of the computation you simply write the assembly code that computes the results. Since you were probably taught to compute only one operation at a time this means that manual computation works on "simple expressions" that exist in a complex expression. Of course converting those simple expressions to assembly is fairly trivial. Therefore anyone who can solve a complex expression by hand can convert it to assembly language following the rules for simple expressions.

9.1.4 Commutative Operators

If "@" represents some operator that operator is commutative if the following relationship is always true:

		(A @ B) = (B @ A)

As you saw in the previous section commutative operators are nice because the order of their operands is immaterial and this lets you rearrange a computation often making that computation easier or more efficient. Often rearranging a computation allows you to use fewer temporary variables. Whenever you encounter a commutative operator in an expression you should always check to see if there is a better sequence you can use to improve the size or speed of your code. The following tables list the commutative and non-commutative operators you typically find in high level languages:

Chapter Nine: Arithmetic And Logical Operations (Part 1)
27 SEP 1996