Lecture 6
Conditional Processing
Assembly Language for
Intel-Based Computers,
4th edition
Kip R. Irvine
1
Outline
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2
Boolean and Comparison Instructions
Conditional Jumps
Conditional Loops
High-Level Logic Structures
Logic Instructions
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3
Syntax for AND, OR, XOR, and TEST:
op-code destination, source
AND, OR and XOR perform the Boolean bitwise
operation and store the result into destination.
TEST is just an AND but the result is not stored
Both operands must be of the same type
 either byte, word or doubleword
Both operands cannot be memory at same time
 again: memory to memory operations are forbidden
They modify OF, CF, SF, ZF, PF according to the
result of the operation (as usual)
Logic Instructions (cont.)
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The source is often an immediate operand
called a bit mask: used to fix certain bits to
0 or 1
To clear a bit we use an AND since:
 0 AND
b = 0 (b is cleared)
 1 AND b = b (b is conserved)
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Ex: to clear the sign bit of AL without
affecting the others, we do:
 AND
al,7Fh ;msb of AL is cleared
since 7Fh = 0111 1111b
4
Logic Instructions (cont.)
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To set (ie: set to 1) certain bits, we use OR:
1
OR b = 1
 0 OR b = b
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To set the sign bit of AH, we do:
 OR
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(b is set)
(b is conserved)
ah,80h
To test if ECX=0 we can do:
 OR
ecx,ecx
 since
this does not change the number in ECX
and set ZF=1 if and only if ecx=0
5
Logic Instructions (cont.)
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XOR can be used to complement,
conserve, or clear certain bits because:
b
XOR 1 = NOT(b) (b is complemented)
 b XOR 0 = b
(b is conserved)
 b XOR b = 0
(b is cleared)
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Ex: to initialize a register to 0 we can use a
two-byte instruction:
 XOR
ax,ax
 instead
of the three-byte instruction:
 MOV ax,0
This was faster on 8088's
6
Logic Instructions (cont.)
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7
Using OR and XOR in strange ways:
XOR ax, ax
MOV ax, 0
OR ax, ax
CMP ax, 0
bytes
2
3
86/88
3 cycles
4 cycles
286
2 cycles
2 cycles
386
2 cycles
2 cycles
486
1 cycle
1 cycle
Machine code
33 C0
0B C0
B8 0000
83 F8 00
Logic Instructions (cont.)
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To convert from upper case to lower case we can
use the usual method:
 ADD dl,20h
Bit 5
But
 OR
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8
"A" =
"a" =
0100 0001b
0110 0001b
dl,20h ;converts from upper to lower case
 AND dl,0DFh ;converts from lower to upper case
since DFh = 1101 1111b
OR dl,20h
and
AND dh, 20h
leaves lower case characters in dl unchanged
and upper case characters in dh unchanged
Logic Instructions (cont.)
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To invert all the bits (ones complement):
NOT destination
 does not affect any flag and destination cannot
be an immediate operand
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Reflection

9
What are the easiest ways to clearing or
setting individual CPU flags?
The CMP instruction
CMP destination,source
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10
Performs: destination - source but does
not store the result of this subtraction into
destination
But the flags are affected like SUB
Same restrictions on operands as SUB
Very often used just before performing a
conditional jump
CMP is similar to TEST except TEST does
AND operation
Conditional Jumps
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Jumps Based On . . .
Specific
flags
Equality
Unsigned
comparisons
Signed Comparisons
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11
Applications
Encrypting a String
Bit Test (BT) Instruction
Jcond Instruction
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A conditional jump instruction branches to a label
when specific register or flag conditions are met
Examples:
JB, JC jump to a label if the Carry flag is set
 JE, JZ jump to a label if the Zero flag is set
 JS jumps to a label if the Sign flag is set
 JNE, JNZ jump to a label if the Zero flag is clear
 JECXZ jumps to a label if ECX equals 0

12
Jumps Based on Specific Flags
13
Jumps Based on Equality
14
Jumps Based on Unsigned
Comparisons
15
Jumps Based on Signed Comparisons
16
Applications (1 of 5)
• Task: Jump to a label if unsigned EAX is greater than EBX
• Solution: Use CMP, followed by JA
cmp eax,ebx
ja Larger
• Task: Jump to a label if signed EAX is greater than EBX
• Solution: Use CMP, followed by JG
cmp eax,ebx
jg Greater
17
Applications (2 of 5)
• Jump to label L1 if unsigned EAX is less than or equal to Val1
cmp eax,Val1
jbe L1
; below or equal
• Jump to label L1 if signed EAX is less than or equal to Val1
cmp eax,Val1
jle L1
18
Applications (3 of 5)
• Compare unsigned AX to BX, and copy the larger of the two
into a variable named Large
mov
cmp
jna
mov
Next:
Large,bx
ax,bx
Next
Large,ax
• Compare signed AX to BX, and copy the smaller of the two
into a variable named Small
mov
cmp
jnl
mov
Next:
19
Small,ax
bx,ax
Next
Small,bx
Applications (4 of 5)
• Jump to label L1 if the memory word pointed to by ESI equals
Zero
cmp WORD PTR [esi],0
je L1
• Jump to label L2 if the doubleword in memory pointed to by
EDI is even
test DWORD PTR [edi],1
jz
L2
20
Applications (5 of 5)
• Task: Jump to label L1 if bits 0, 1, and 3 in AL are all set.
• Solution: Clear all bits except bits 0, 1,and 3. Then
compare the result with 00001011 binary.
and al,00001011b
cmp al,00001011b
je L1
21
; clear unwanted bits
; check remaining bits
; all set? jump to L1
Your turn . . .
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22
Write code that jumps to label L1 if either
bit 4, 5, or 6 is set in the BL register.
Write code that jumps to label L1 if bits 4,
5, and 6 are all set in the BL register.
Write code that jumps to label L3 if EAX is
negative.
Encrypting a String
The following loop uses the XOR instruction to transform every
character in a string into a new value.
KEY = 239
; can be any byte value
BUFMAX = 128
.data
buffer BYTE BUFMAX+1 DUP(0)
bufSize DWORD BUFMAX
.code
mov ecx,bufSize
mov esi,0
L1:
xor buffer[esi],KEY
inc esi
loop L1
23
; loop counter
; index 0 in buffer
; translate a byte
; point to next byte
String Encryption Program
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Tasks:
 Input
a message (string) from the user
 Encrypt the message
 Display the encrypted message
 Decrypt the message
 Display the decrypted message
View the Encrypt.asm program's source code. Sample output:
Enter the plain text: Attack at dawn.
Cipher text: «¢¢Äîä-Ä¢-ïÄÿü-Gs
Decrypted: Attack at dawn.
24
Conditional Loop Instructions
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25
LOOPZ and LOOPE
LOOPNZ and LOOPNE
LOOPZ and LOOPE
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Syntax:
LOOPE destination
LOOPZ destination
Logic:
 ECX  ECX – 1
 if ECX > 0 and ZF=1, jump to destination
Useful when scanning an array for the first
element that does not match a given value.
LOOPNZ and LOOPNE
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LOOPNZ (LOOPNE) is a conditional loop
instruction
Syntax:
LOOPNZ destination
LOOPNE destination
Logic:
ECX  ECX – 1;
 if ECX > 0 and ZF=0, jump to destination

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Useful when scanning an array for the first
element that matches a given value.
LOOPNZ Example
The following code finds the first positive value in an array:
.data
array SWORD -3,-6,-1,-10,10,30,40,4
sentinel SWORD 0
.code
mov esi,OFFSET array
mov ecx,LENGTHOF array
next:
test WORD PTR [esi],8000h ; test sign bit
pushfd
; push flags on stack
add esi,TYPE array
popfd
; pop flags from stack
loopnz next
; continue loop
jnz quit
; none found
sub esi,TYPE array
; ESI points to value
quit:
28
Your turn . . .
Locate the first nonzero value in the array. If none is found, let
ESI point to the sentinel value:
.data
array SWORD 50 DUP(?)
sentinel SWORD 0FFFFh
.code
mov esi,OFFSET array
mov ecx,LENGTHOF array
L1: cmp WORD PTR [esi],0
(fill in your code here)
quit:
29
; check for zero
. . . (solution)
.data
array SWORD 50 DUP(?)
sentinel SWORD 0FFFFh
.code
mov esi,OFFSET array
mov ecx,LENGTHOF array
L1: cmp WORD PTR [esi],0
pushfd
add esi,TYPE array
popfd
loope L1
jz quit
sub esi,TYPE array
quit:
30
; check for zero
; push flags on stack
;
;
;
;
pop flags from stack
continue loop
none found
ESI points to value
Conditional Structures
31
•
Block-Structured IF Statements
•
Compound Expressions with AND
•
Compound Expressions with OR
•
WHILE Loops
•
Table-Driven Selection
Block-Structured IF Statements
Assembly language programmers can easily
translate logical statements written in C++/Java into
assembly language. For example:
if( op1 == op2 )
X = 1;
else
X = 2;
32
mov
cmp
jne
mov
jmp
L1: mov
L2:
eax,op1
eax,op2
L1
X,1
L2
X,2
Your turn . . .
Implement the following pseudocode in
assembly language. All values are unsigned:
if( ebx <= ecx )
{
eax = 5;
edx = 6;
}
cmp
ja
mov
mov
next:
ebx,ecx
next
eax,5
edx,6
(There are multiple correct solutions to this problem.)
33
Your turn . . .
Implement the following pseudocode in assembly
language. All values are 32-bit signed integers:
if( var1
var3 =
else
{
var3 =
var4 =
}
<= var2 )
10;
6;
7;
mov
cmp
jle
mov
mov
jmp
L1: mov
L2:
eax,var1
eax,var2
L1
var3,6
var4,7
L2
var3,10
(There are multiple correct solutions to this problem.)
34
Compound Expression with AND (1 of 3)
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When implementing the logical AND operator,
consider that HLLs use short-circuit evaluation
In the following example, if the first expression is
false, the second expression is skipped:
if (al > bl) AND (bl > cl)
X = 1;
35
Compound Expression with AND (2 of 3)
if (al > bl) AND (bl > cl)
X = 1;
This is one possible implementation . . .
cmp al,bl
ja L1
jmp next
; first expression...
cmp bl,cl
ja L2
jmp next
; second expression...
L1:
L2:
mov X,1
next:
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; both are true
; set X to 1
Compound Expression with AND (3 of 3)
if (al > bl) AND (bl > cl)
X = 1;
But the following implementation uses 29% less code by
reversing the first relational operator. We allow the program to
"fall through" to the second expression:
cmp
jbe
cmp
jbe
mov
next:
37
al,bl
next
bl,cl
next
X,1
;
;
;
;
;
first expression...
quit if false
second expression...
quit if false
both are true
Your turn . . .
Implement the following pseudocode in
assembly language. All values are unsigned:
if( ebx <= ecx
&& ecx > edx )
{
eax = 5;
edx = 6;
}
cmp
ja
cmp
jbe
mov
mov
next:
ebx,ecx
next
ecx,edx
next
eax,5
edx,6
(There are multiple correct solutions to this problem.)
38
Compound Expression with OR (1 of 2)
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When implementing the logical OR operator,
consider that HLLs use short-circuit evaluation
In the following example, if the first expression is
true, the second expression is skipped:
if (al > bl) OR (bl > cl)
X = 1;
39
Compound Expression with OR (1 of 2)
if (al > bl) OR (bl > cl)
X = 1;
We can use "fall-through" logic to keep the code as short as
possible:
cmp
ja
cmp
jbe
L1: mov
next:
40
al,bl
L1
bl,cl
next
X,1
;
;
;
;
;
is AL > BL?
yes
no: is BL > CL?
no: skip next statement
set X to 1
WHILE Loops
A WHILE loop is really an IF statement followed by the body
of the loop, followed by an unconditional jump to the top of
the loop. Consider the following example:
while( eax < ebx)
eax = eax + 1;
This is a possible implementation:
top: cmp
jae
inc
jmp
next:
41
eax,ebx
next
eax
top
;
;
;
;
check loop condition
false? exit loop
body of loop
repeat the loop
Your turn . . .
Implement the following loop, using unsigned 32-bit integers:
while( ebx <= val1)
{
ebx = ebx + 5;
val1 = val1 - 1
}
top: cmp
ja
add
dec
jmp
next:
42
ebx,val1
next
ebx,5
val1
top
; check loop condition
; false? exit loop
; body of loop
; repeat the loop
Using the .IF Directive
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43
Runtime Expressions
Relational and Logical
Operators
MASM-Generated Code
.REPEAT Directive
.WHILE Directive
Runtime Expressions
• .IF, .ELSE, .ELSEIF, and .ENDIF can be used to evaluate
runtime expressions and create block-structured IF
statements.
• Examples:
.IF eax > ebx
mov edx,1
.ELSE
mov edx,2
.ENDIF
.IF eax > ebx && eax > ecx
mov edx,1
.ELSE
mov edx,2
.ENDIF
• MASM generates "hidden" code for you, consisting of
code labels, CMP and conditional jump instructions.
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Relational and Logical Operators
45
MASM-Generated Code
.data
val1
DWORD 5
result DWORD ?
.code
mov eax,6
.IF eax > val1
mov result,1
.ENDIF
Generated code:
mov eax,6
cmp eax,val1
jbe @C0001
mov result,1
@C0001:
MASM automatically generates an unsigned jump (JBE)
because val1 is unsigned.
46
MASM-Generated Code
.data
val1
SDWORD 5
result SDWORD ?
.code
mov eax,6
.IF eax > val1
mov result,1
.ENDIF
Generated code:
mov eax,6
cmp eax,val1
jle @C0001
mov result,1
@C0001:
MASM automatically generates a signed jump (JLE) because
val1 is signed.
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MASM-Generated Code
.data
result DWORD ?
.code
mov ebx,5
mov eax,6
.IF eax > ebx
mov result,1
.ENDIF
Generated code:
mov ebx,5
mov eax,6
cmp eax,ebx
jbe @C0001
mov result,1
@C0001:
MASM automatically generates an unsigned jump (JBE) when
both operands are registers . . .
48
MASM-Generated Code
.data
result SDWORD ?
.code
mov ebx,5
mov eax,6
.IF SDWORD PTR eax > ebx
mov result,1
.ENDIF
Generated code:
mov ebx,5
mov eax,6
cmp eax,ebx
jle @C0001
mov result,1
@C0001:
. . . unless you prefix one of the register operands with the
SDWORD PTR operator. Then a signed jump is generated.
49
.REPEAT Directive
Executes the loop body before testing the loop condition
associated with the .UNTIL directive.
Example:
; Display integers 1 – 10:
mov eax,0
.REPEAT
inc eax
call WriteDec
call Crlf
.UNTIL eax == 10
50
.WHILE Directive
Tests the loop condition before executing the loop body The
.ENDW directive marks the end of the loop.
Example:
; Display integers 1 – 10:
mov eax,0
.WHILE eax < 10
inc eax
call WriteDec
call Crlf
.ENDW
51
Macros
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52
Introducing Macros
Defining Macros
Invoking Macros
Macro Examples
Nested Macros
Example Program: Wrappers
Introducing Macros
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A macro1 is a named block of assembly language
statements.
Once defined, it can be invoked (called) one or
more times.
During the assembler's preprocessing step, each
macro call is expanded into a copy of the macro.
The expanded code is passed to the assembly
step, where it is checked for correctness.
1Also
53
called a macro procedure.
Defining Macros
•
A macro must be defined before it can be used.
•
Parameters are optional.
•
•
Each parameter follows the rules for identifiers.
It is a string that is assigned a value when the
macro is invoked.
Syntax:
macroname MACRO [parameter-1, parameter-2,...]
statement-list
ENDM
54
mNewLine Macro Example
This is how you define and invoke a simple macro.
mNewLine MACRO
call Crlf
ENDM
.data
.code
mNewLine
; define the macro
; invoke the macro
The assembler will substitute "call crlf" for "mNewLine".
55
mPutChar Macro
Writes a single character to standard output.
Definition:
Invocation:
Expansion:
56
mPutchar MACRO char
push eax
mov al,char
call WriteChar
pop eax
ENDM
.code
mPutchar 'A'
1
1
1
1
push eax
mov al,'A'
call WriteChar
pop eax
viewed in the
listing file
Invoking Macros (1 of 2)
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57
When you invoke a macro, each argument you
pass matches a declared parameter.
Each parameter is replaced by its corresponding
argument when the macro is expanded.
When a macro expands, it generates assembly
language source code.
Arguments are treated as simple text by the
preprocessor.
Invoking Macros (2 of 2)
Relationships between macros, arguments, and parameters:
macro
invocation
statement
passes
text
consists
of
assembly
code
argument
generates
replaces
parameter
58
declared
inside
macro
mWriteStr Macro (1 of 2)
Provides a convenient way to display a string, by
passing the string name as an argument.
mWriteStr MACRO buffer
push edx
mov edx,OFFSET buffer
call WriteString
pop edx
ENDM
.data
str1 BYTE "Welcome!",0
.code
mWriteStr str1
59
mWriteStr Macro (2 of 2)
The expanded code shows how the str1 argument
replaced the parameter named buffer:
60
mWriteStr
push
mov
call
pop
ENDM
MACRO buffer
edx
edx,OFFSET buffer
WriteString
edx
1
1
1
1
edx
edx,OFFSET str1
WriteString
edx
push
mov
call
pop
Invalid Argument


If you pass an invalid argument, the error is
caught when the expanded code is assembled.
Example: mPutchar MACRO char
push eax
mov al,char
call WriteChar
pop eax
ENDM
.code
mPutchar 1234h
61
1
1
1
1
push eax
mov al,1234h
call WriteChar
pop eax
; error!
Blank Argument


If you pass a blank argument, the error is also
caught when the expanded code is assembled.
Example: mPutchar MACRO char
push eax
mov al,char
call WriteChar
pop eax
ENDM
.code
mPutchar
62
1
1
1
1
push eax
mov al,
call WriteChar
pop eax
Your turn . . .

Write a macro named mMove32 that receives
two 32-bit memory operands. The macro should
move the source operand to the destination
operand.
. . . Solution
mMove32 MACRO destination,source
push eax
mov eax,source
mov destination,eax
pop eax
ENDM
63