Computer Organization &
Computer Organization &
Assembly Languages Assembly Languages
Pu-Jen Cheng
Data Transfers, Addressing & Arithmetic
Adapted from the slides prepared by Kip Irvine for the book, Assembly Language for Intel-Based Computers, 5th Ed.
Chapter Overview
Data Transfer Instructions
Addition and Subtraction
Data-Related Operators and Directives
Indirect Addressing
JMP and LOOP Instructions
Data Transfer Instructions
Operand Types
Instruction Operand Notation
Direct Memory Operands
MOV Instruction
Zero & Sign Extension
XCHG Instruction
Direct-Offset Instructions
Operand Types
Three basic types of operands:
¾ Immediate – a constant integer (8, 16, or 32 bits)
value is encoded within the instruction
¾ Register – the name of a register
register name is converted to a number and
register name is converted to a number and encoded within the instruction
¾ Memory – reference to a location in memory
memory address is encoded within the
instruction, or a register holds the address of a memory location
Instruction Operand Notation
Direct Memory Operands
A direct memory operand is a named reference to storage in memory
The named reference (label) is automatically dereferenced by the assembler
.data
var1 BYTE 10h .code
mov al,var1 ; AL = 10h
mov al,[var1] ; AL = 10h
alternate format
MOV Instruction
• Move from source to destination. Syntax:
MOV destination, source
• Source and destination have the same size
• No more than one memory operand permitted
• CS, EIP, and IP cannot be the destination
.data
count BYTE 100 wVal WORD 2 .code
mov bl,count mov ax,wVal mov count,al
mov al,wVal ; error
mov ax,count ; error
mov eax,count ; error
• No immediate to segment moves
Your Turn . . .
.data
bVal BYTE 100 bVal2 BYTE ?
Explain why each of the following MOV statements are invalid:
wVal WORD 2 dVal DWORD 5 .code
mov ds,45 mov esi,wVal mov eip,dVal mov 25,bVal mov bVal2,bVal
immediate move to DS not permitted size mismatch
EIP cannot be the destination
immediate value cannot be destination memory-to-memory move not permitted
Memory to Memory
.data
var1 WORD ? var2 WORD ? .code
1
mov ax, var1
mov var2, ax
Copy Smaller to Larger
.data
count WORD 1 .code
mov ecx, 0
mov cx, count .data
signedVal SWORD -16 ; FFF0h .code
mov ecx, 0 ; mov ecx, 0FFFFFFFFh mov cx, signedVal
Zero Extension
When you copy a smaller value into a larger destination, the MOVZX instruction fills (extends) the upper half of the
destination with zeros.
1 0 0 0 1 1 1 1 Source 0
mov bl,10001111b
movzx ax,bl ; zero-extension
1 0 0 0 1 1 1 1 Destination 0 0 0 0 0 0 0 0
The destination must be a register.
Sign Extension
The MOVSX instruction fills the upper half of the destination with a copy of the source operand's sign bit.
1 0 0 0 1 1 1 1 Source
mov bl,10001111b
movsx ax,bl ; sign extension
1 0 0 0 1 1 1 1 Destination 1 1 1 1 1 1 1 1
The destination must be a register.
MOVZX & MOVSX
From a smaller location to a larger one mov bx, 0A69Bh
movzx eax, bx ; EAX=0000A69Bh
d bl EDX 0000009Bh
movzx edx, bl ; EDX=0000009Bh movzx cx, bl ; EAX=009Bh
mov bx, 0A69Bh
movsx eax, bx ; EAX=FFFFA69Bh
movsx edx, bl ; EDX=FFFFFF9Bh
movsx cx, bl ; EAX=FF9Bh
LAHF & SAHF
.data
saveflags BYTE ? .code
lahf
mov saveflags, ah ...
mov ah, saveflags
sahf
XCHG Instruction
.data
var1 WORD 1000h
2 2000h
XCHG exchanges the values of two operands. At least one operand must be a register. No immediate operands are permitted.
var2 WORD 2000h .code
xchg ax,bx ; exchange 16-bit regs xchg ah,al ; exchange 8-bit regs xchg var1,bx ; exchange mem, reg xchg eax,ebx ; exchange 32-bit regs
xchg var1,var2 ; error: two memory operands
Direct-Offset Operands
.data
arrayB BYTE 10h,20h,30h,40h .code
A constant offset is added to a data label to produce an effective address (EA). The address is dereferenced to get the value inside its memory location.
mov al,arrayB+1 ; AL = 20h
mov al,[arrayB+1] ; alternative notation
Q: Why doesn't arrayB+1 produce 11h?
Direct-Offset Operands (cont.)
.data
arrayW WORD 1000h,2000h,3000h arrayD DWORD 1,2,3,4
A constant offset is added to a data label to produce an effective address (EA). The address is dereferenced to get the value inside its memory location.
.code
mov ax,[arrayW+2] ; AX = 2000h mov ax,[arrayW+4] ; AX = 3000h
mov eax,[arrayD+4] ; EAX = 00000002h
; Will the following statements assemble?
mov ax,[arrayW-2] ; ??
mov eax,[arrayD+16] ; ??
What will happen when they run?
Your Turn. . .
Write a program that rearranges the values of three doubleword values in the following array as: 3, 1, 2.
.data
arrayD DWORD 1,2,3
• Step1: copy the first value into EAX and exchange it with the value in the second position.
• Step 2: Exchange EAX with the third array value and copy the value in EAX to the first array position.
value in the second position.
mov eax,arrayD
xchg eax,[arrayD+4]
xchg eax,[arrayD+8]
mov arrayD,eax
Evaluate This . . .
• We want to write a program that adds the following three bytes:
.data
myBytes BYTE 80h,66h,0A5h
• What is your evaluation of the following code?
mov al,myBytes
add al,[myBytes+1]
add al [myBytes+2]
add al,[myBytes+2]
• What is your evaluation of the following code?
mov ax,myBytes
add ax,[myBytes+1]
add ax,[myBytes+2]
• Any other possibilities?
Evaluate This . . . (cont)
.data
myBytes BYTE 80h,66h,0A5h
• How about the following code. Is anything missing?
movzx ax,myBytes
mov bl,[myBytes+1]
add ax,bx add ax,bx
mov bl,[myBytes+2]
add ax,bx ; AX = sum
Yes: Move zero to BX before the MOVZX instruction.
What's Next
Data Transfer Instructions
Addition and Subtraction
Data-Related Operators and Directives
Indirect Addressing
JMP and LOOP Instructions
Addition and Subtraction
INC and DEC Instructions
ADD and SUB Instructions
NEG Instruction
Implementing Arithmetic Expressions
Flags Affected by Arithmetic
¾ Zero
¾ Sign
¾ Carry
¾ Overflow
INC and DEC Instructions
Add 1, subtract 1 from destination operand
¾ operand may be register or memory
INC
destination¾ Logic: destination ← destination + 1
DEC
destination¾ Logic: destination ← destination – 1
INC and DEC Examples
.data
myWord WORD 1000h
myDword DWORD 10000000h .code
inc myWord ; 1001h
dec myWord ; 1000h
inc myDword ; 10000001h mov ax,00FFh
inc ax ; AX = 0100h
mov ax,00FFh
inc al ; AX = 0000h
Your Turn...
Show the value of the destination operand after each of the following instructions executes:
.data
myByte BYTE 0FFh, 0 .code
mov al,myByte ; AL = mov ah,[myByte+1] ; AH =
dec ah ; AH =
inc al ; AL =
dec ax ; AX =
FFh 00h FFh 00h FEFF
ADD and SUB Instructions
• ADD destination, source
• Logic: destination ← destination + source
• SUB destination, source
• Logic: destination ← destination – source
• Same operand rules as for the MOV instructionSa e ope a d u es as o e O s uc o
ADD and SUB Examples
.data
var1 DWORD 10000h var2 DWORD 20000h
.code ; ---EAX---
mov eax,var1 ; 00010000h add eax,var2 ; 00030000h add ax,0FFFFh ; 0003FFFFh
add eax,1 ; 00040000h
sub ax,1 ; 0004FFFFh
NEG (negate) Instruction
.data
valB BYTE -1
valW SWORD +32767 d
Reverses the sign of an operand. Operand can be a register or memory operand.
.code
mov al,valB ; AL = -1
neg al ; AL = +1
neg valW ; valW = -32767
Suppose AX contains –32,768 and we apply NEG to it.
Will the result be valid?
Implementing Arithmetic Expressions
Rval SDWORD ? X l SDWORD 26
HLL compilers translate mathematical expressions into assembly language. You can do it also. For example:
Rval = -Xval + (Yval – Zval)
Xval SDWORD 26 Yval SDWORD 30 Zval SDWORD 40 .code
mov eax,Xval
neg eax ; EAX = -26
mov ebx,Yval
sub ebx,Zval ; EBX = -10 add eax,ebx
mov Rval,eax ; -36
Your Turn...
Translate the following expression into assembly language.
Do not permit Xval, Yval, or Zval to be modified:
Rval = Xval - (-Yval + Zval)
Assume that all values are signed doublewords.
mov ebx,Yval neg ebx
add ebx,Zval mov eax,Xval sub eax,ebx mov Rval,eax
Flags Affected by Arithmetic
The ALU has a number of status flags that
reflect the outcome of arithmetic (and bitwise) operations
¾ based on the contents of the destination operand
Essential flags:
Essential flags:
¾ Zero flag – set when destination equals zero
¾ Sign flag – set when destination is negative
¾ Carry flag – set when unsigned value is out of range
¾ Overflow flag – set when signed value is out of range
The MOV instruction never affects the flags.
Concept Map
ALU conditional jumps
ith ti & bit i
part of
CPU
executes executes
status flags
branching logic arithmetic & bitwise
operations attached to used by provide
affect
You can use diagrams such as these to express the relationships between assembly language concepts.
Zero Flag (ZF)
mov cx,1
sub cx,1 ; CX = 0, ZF = 1 mov ax,0FFFFh
inc ax ; AX = 0, ZF = 1
The Zero flag is set when the result of an operation produces zero in the destination operand.
inc ax ; AX = 1, ZF = 0
Remember...
• A flag is set when it equals 1.
• A flag is clear when it equals 0.
Sign Flag (SF)
mov cx,0
sub cx,1 ; CX = -1, SF = 1
add cx,2 ; CX = 1, SF = 0
The Sign flag is set when the destination operand is
negative. The flag is clear when the destination is positive.
The sign flag is a copy of the destination's highest bit:
mov al,0
sub al,1 ; AL = 11111111b, SF = 1 add al,2 ; AL = 00000001b, SF = 0
Signed and Unsigned Integers A Hardware Viewpoint
All CPU instructions operate exactly the same on signed and unsigned integers
The CPU cannot distinguish between signed and unsigned integers
YOU the programmer are solely responsible for
YOU, the programmer, are solely responsible for
using the correct data type with each instruction
Overflow and Carry Flags A Hardware Viewpoint
How the ADD instruction modifies OF and CF:
¾ OF = (carry out of the MSB) XOR (carry into the MSB)
¾ CF = (carry out of the MSB)
How the SUB instruction modifies OF and CF:
¾ NEG the source and ADD it to the destination
¾ OF = (carry out of the MSB) XOR (carry into the MSB)
¾ CF = INVERT (carry out of the MSB)
MSB = Most Significant Bit (high-order bit) XOR = eXclusive-OR operation
NEG = Negate (same as SUB 0,operand )
Carry Flag (CF)
The Carry flag is set when the result of an operation
generates an unsigned value that is out of range (too big or too small for the destination operand).
mov al,0FFh
add al,1 ; CF = 1, AL = 00
; Try to go below zero:
mov al,0
sub al,1 ; CF = 1, AL = FF
In the second example, we tried to generate a negative value.
Unsigned values cannot be negative, so the Carry flag signaled an error condition.
Your Turn . . .
mov ax,00FFh
add ax,1 ; AX= SF= ZF= CF=
sub ax,1 ; AX= SF= ZF= CF=
For each of the following marked entries, show the values of the destination operand and the Sign, Zero, and Carry flags:
0100h 0 0 0 00FFh 0 0 0 add al,1 ; AL= SF= ZF= CF=
mov bh,6Ch
add bh,95h ; BH= SF= ZF= CF=
mov al,2
sub al,3 ; AL= SF= ZF= CF=
00h 0 1 1 01h 0 0 1
FFh 1 0 1
Overflow Flag (OF)
The Overflow flag is set when the signed result of an operation is invalid or out of range.
; Example 1 mov al,+127
add al,1 ; OF = 1, AL = ??
; Example 2 mov al,7Fh
add al,1 ; OF = 1, AL = 80h
The two examples are identical at the binary level
because 7Fh equals +127. To determine the value of the destination operand, it is often easier to calculate in
hexadecimal.
A Rule of Thumb
When adding two integers, remember that the Overflow flag is only set when . . .
¾ Two positive operands are added and their sum is negative
¾ Two negative operands are added and their sum is g positive
What will be the values of the Overflow flag?
mov al,80h
add al,92h ; OF =
mov al,-2
add al,+127 ; OF =
1
0
Your Turn . . .
mov al,-128
neg al ; CF = OF =
mov ax,8000h
What will be the values of the given flags after each operation?
0 1
add ax,2 ; CF = OF =
mov ax,0
sub ax,2 ; CF = OF =
mov al,-5
sub al,+125 ; CF = OF = 0 0
1 0
0 1
What's Next
Data Transfer Instructions
Addition and Subtraction
Data-Related Operators and Directives
Indirect Addressing
JMP and LOOP Instructions
Data-Related Operators and Directives
OFFSET Operator
PTR Operator
TYPE Operator
LENGTHOF Operator
SIZEOF Operator
LABEL Directive
OFFSET Operator
OFFSET returns
the distance in bytes, of a label from the beginning of its enclosing segment
¾ Protected mode: 32 bits
¾ Real mode: 16 bits
offset
myByte data segment:
The Protected-mode programs we write only have a single segment (we use the flat memory model).
OFFSET Examples
.data
bVal BYTE ? wVal WORD ? dVal DWORD ?
Let's assume that the data segment begins at 00404000h:
dVal2 DWORD ? .code
mov esi,OFFSET bVal ; ESI = 00404000 mov esi,OFFSET wVal ; ESI = 00404001 mov esi,OFFSET dVal ; ESI = 00404003 mov esi,OFFSET dVal2 ; ESI = 00404007
Relating to C/C++
; C++ version:
h [ ]
The value returned by OFFSET is a pointer. Compare the following code written for both C++ and assembly language:
char array[1000];
char * p = array;
.data
array BYTE 1000 DUP(?) .code
mov esi,OFFSET array ; ESI is p
PTR Operator
.data
myDouble DWORD 12345678h .code
Overrides the default type of a label (variable).
Provides the flexibility to access part of a variable.
mov ax,myDouble ; error – why?
mov ax,WORD PTR myDouble ; loads 5678h mov WORD PTR myDouble,4321h ; saves 4321h
Recall that little endian order is used when storing data in memory (see Section 3.4.9).
Little Endian Order
Little endian order refers to the way Intel stores integers in memory.
Multi-byte integers are stored in reverse order, with the least significant byte stored at the lowest address
For example, the doubleword 12345678h would be stored as:
78 0000
56 34 12
0001 0002 0003
offset
byte When integers are loaded from memory into registers, the bytes are automatically re-reversed into their correct positions.
PTR Operator Examples
.data
myDouble DWORD 12345678h
12345678 5678 0000 1234
78 56 34
0001 0002 offset doubleword word byte
myDouble myDouble + 1
D bl 2
1234 34 12
0002 0003
myDouble + 2 myDouble + 3
mov al,BYTE PTR myDouble ; AL = 78h mov al,BYTE PTR [myDouble+1] ; AL = 56h mov al,BYTE PTR [myDouble+2] ; AL = 34h mov ax,WORD PTR myDouble ; AX = 5678h mov ax,WORD PTR [myDouble+2] ; AX = 1234h
PTR Operator (cont.)
.data
myBytes BYTE 12h,34h,56h,78h
PTR can also be used to combine elements of a
smaller data type and move them into a larger operand.
The CPU will automatically reverse the bytes.
.code
mov ax,WORD PTR [myBytes] ; AX = 3412h mov ax,WORD PTR [myBytes+2] ; AX = 7856h
mov eax,DWORD PTR myBytes ; EAX = 78563412h
Your Turn . . .
.data
varB BYTE 65h,31h,02h,05h varW WORD 6543h,1202h
varD DWORD 12345678h
Write down the value of each destination operand:
.code
mov ax,WORD PTR [varB+2] ; a.
mov bl,BYTE PTR varD ; b.
mov bl,BYTE PTR [varW+2] ; c.
mov ax,WORD PTR [varD+2] ; d.
mov eax,DWORD PTR varW ; e.
0502h 78h 02h 1234h
12026543h
TYPE Operator
The TYPE operator returns the size, in bytes, of a single element of a data declaration.
.data
var1 BYTE ? var2 WORD ? var3 DWORD ? var4 QWORD ? .code
mov eax,TYPE var1 ; 1 mov eax,TYPE var2 ; 2 mov eax,TYPE var3 ; 4 mov eax,TYPE var4 ; 8
LENGTHOF Operator
.data LENGTHOF
byte1 BYTE 10,20,30 ; 3
array1 WORD 30 DUP(?),0,0 ; 32 2 WORD 5 DUP(3 DUP(?)) 15
The LENGTHOF operator counts the number of elements in a single data declaration.
array2 WORD 5 DUP(3 DUP(?)) ; 15
array3 DWORD 1,2,3,4 ; 4
digitStr BYTE "12345678",0 ; 9 .code
mov ecx,LENGTHOF array1 ; 32
SIZEOF Operator
.data SIZEOF
byte1 BYTE 10,20,30 ; 3
array1 WORD 30 DUP(?),0,0 ; 64 2 WORD 5 DUP(3 DUP(?)) 30
The SIZEOF operator returns a value that is equivalent to multiplying LENGTHOF by TYPE.
array2 WORD 5 DUP(3 DUP(?)) ; 30
array3 DWORD 1,2,3,4 ; 16
digitStr BYTE "12345678",0 ; 9 .code
mov ecx,SIZEOF array1 ; 64
Spanning Multiple Lines
.data
A data declaration spans multiple lines if each line
(except the last) ends with a comma. The LENGTHOF and SIZEOF operators include all lines belonging to the declaration:
.data
array WORD 10,20, 30,40,
50,60 .code
mov eax,LENGTHOF array ; 6 mov ebx,SIZEOF array ; 12
Spanning Multiple Lines (cont.)
data
In the following example, array identifies only the first WORD declaration. Compare the values returned by LENGTHOF and SIZEOF here to those in the
previous slide:
.data
array WORD 10,20 WORD 30,40 WORD 50,60 .code
mov eax,LENGTHOF array ; 2 mov ebx,SIZEOF array ; 4
LABEL Directive
Assigns an alternate label name and type to an existing storage location
LABEL does not allocate any storage of its own
Removes the need for the PTR operator
data .data
dwList LABEL DWORD wordList LABEL WORD
intList BYTE 00h,10h,00h,20h .code
mov eax,dwList ; 20001000h mov cx,wordList ; 1000h
mov dl,intList ; 00h
What's Next
Data Transfer Instructions
Addition and Subtraction
Data-Related Operators and Directives
Indirect Addressing
JMP and LOOP Instructions
Indirect Addressing
Indirect Operands
Array Sum Example
Indexed Operands
Pointers
Indirect Operands
.data
val1 BYTE 10h,20h,30h code
An indirect operand holds the address of a variable, usually an array or string. It can be dereferenced (just like a pointer).
.code
mov esi,OFFSET val1
mov al,[esi] ; dereference ESI (AL = 10h) inc esi
mov al,[esi] ; AL = 20h
inc esi
mov al,[esi] ; AL = 30h
Indirect Operands (cont.)
.data
myCount WORD 0 .code
Use PTR to clarify the size attribute of a memory operand.
mov esi,OFFSET myCount
inc [esi] ; error: ambiguous
inc WORD PTR [esi] ; ok
Should PTR be used here?
add [esi],20
yes, because [esi] could point to a byte, word, or doubleword
Array Sum Example
.data
arrayW WORD 1000h,2000h,3000h .code
Indirect operands are ideal for traversing an array. Note that the register in brackets must be incremented by a value that matches the array type.
mov esi,OFFSET arrayW mov ax,[esi]
add esi,2 ; or: add esi,TYPE arrayW add ax,[esi]
add esi,2
add ax,[esi] ; AX = sum of the array
ToDo: Modify this example for an array of doublewords.
Indexed Operands
.data
arrayW WORD 1000h,2000h,3000h code
An indexed operand adds a constant to a register to
generate an effective address. There are two notational forms: [label + reg] label[reg]
.code
mov esi,0
mov ax,[arrayW + esi] ; AX = 1000h
mov ax,arrayW[esi] ; alternate format add esi,2
add ax,[arrayW + esi]
etc.
ToDo: Modify this example for an array of doublewords.
Index Scaling
.data
arrayB BYTE 0,1,2,3,4,5 arrayW WORD 0,1,2,3,4,5
You can scale an indirect or indexed operand to the offset of an array element. This is done by multiplying the index by the array's TYPE:
arrayD DWORD 0,1,2,3,4,5 .code
mov esi,4
mov al,arrayB[esi*TYPE arrayB] ; 04 mov bx,arrayW[esi*TYPE arrayW] ; 0004
mov edx,arrayD[esi*TYPE arrayD] ; 00000004
Pointers
.data
arrayW WORD 1000h,2000h,3000h ptrW DWORD arrayW
d
You can declare a pointer variable that contains the offset of another variable.
.code
mov esi,ptrW
mov ax,[esi] ; AX = 1000h
Alternate format:
ptrW DWORD OFFSET arrayW
What's Next
Data Transfer Instructions
Addition and Subtraction
Data-Related Operators and Directives
Indirect Addressing
JMP and LOOP Instructions
JMP and LOOP Instructions
JMP Instruction
LOOP Instruction
LOOP Example
Summing an Integer Array
Copying a String
JMP Instruction
• JMP is an unconditional jump to a label that is usually within the same procedure.
• Syntax: JMP target
• Logic: EIP ← target
top:
. .
jmp top
• Example:
A jump outside the current procedure must be to a
special type of label called a global label (see Section 5.5.2 for details).
LOOP Instruction
• The LOOP instruction creates a counting loop
• Syntax: LOOP target
• Logic:
• ECX ← ECX – 1
• if ECX != 0, jump to target
• Implementation:
• The assembler calculates the distance, in bytes, between the offset of the following instruction and the offset of the target label. It is called the relative offset.
• The relative offset is added to EIP.
LOOP Example
00000000 66 B8 0000 mov ax,0 00000004 B9 00000005 mov ecx,5
The following loop calculates the sum of the integers 5 + 4 + 3 +2 + 1:
offset machine code source code
00000009 66 03 C1 L1: add ax,cx
0000000C E2 FB loop L1
0000000E
When LOOP is assembled, the current location = 0000000E (offset of the next instruction). –5 (FBh) is added to the the current location, causing a jump to location 00000009:
00000009 ← 0000000E + FB
Your Turn . . .
If the relative offset is encoded in a single signed byte, (a) what is the largest possible backward jump?
(b) what is the largest possible forward jump?
(a) −128 (b) +127
Your Turn . . .
What will be the final value of AX?
mov ax,6 mov ecx,4 L1:
inc ax loop L1
10
How many times will the loop execute?
mov ecx,0 X2:
inc ax loop X2
4,294,967,296
Nested Loop
If you need to code a loop within a loop, you must save the outer loop counter's ECX value.
.data
count DWORD ? .code
mov ecx,100 ; set outer loop count L1:
mov count,ecx ; save outer loop count mov ecx,20 ; set inner loop count L2: .
.
loop L2 ; repeat the inner loop mov ecx,count ; restore outer loop count loop L1 ; repeat the outer loop
Summing an Integer Array
.data
intarray WORD 100h,200h,300h,400h .code
The following code calculates the sum of an array of 16-bit integers.
mov edi,OFFSET intarray ; address of intarray mov ecx,LENGTHOF intarray ; loop counter
mov ax,0 ; zero the accumulator
L1:
add ax,[edi] ; add an integer
add edi,TYPE intarray ; point to next integer
loop L1 ; repeat until ECX = 0
Your Turn . . .
What changes would you make to the program on the previous slide if you were summing a doubleword array?
Copying a String
.data
source BYTE "This is the source string",0 target BYTE SIZEOF source DUP(0)
d
good use of SIZEOF
The following code copies a string from source to target:
.code
mov esi,0 ; index register
mov ecx,SIZEOF source ; loop counter L1:
mov al,source[esi] ; get char from source mov target[esi],al ; store it in the target
inc esi ; move to next character
loop L1 ; repeat for entire string
Your Turn . . .
Rewrite the program shown in the previous slide, using indirect addressing rather than indexed addressing.
CMP Instruction (See 6.2.7)
Compares the destination operand to the source operand (both are unsigned)
Syntax: CMP destination, source
Example: destination == source
mov al,5
cmp al,5 ; Zero flag set
• Example: destination < source
mov al,4
cmp al,5 ; Carry flag set
CMP Instruction
Example: destination > source
mov al,6
cmp al,5 ; ZF = 0, CF = 0
(b th th Z d C fl l )
(both the Zero and Carry flags are clear)
J cond Instruction
A conditional jump instruction branches to a label when specific register or flag conditions are met
Examples:
JB JC j t l b l if th C fl i t
¾ JB, JC jump to a label if the Carry flag is set
¾ JE, JZ jump to a label if the Zero flag is set
¾ JS jumps to a label if the Sign flag is set
¾ JNE, JNZ jump to a label if the Zero flag is clear
¾ JECXZ jumps to a label if ECX equals 0
J cond Ranges
Prior to the 386:
¾ jump must be within –128 to +127 bytes from current location counter
IA-32 processors:
32 bit offset permits jump anywhere in memory
¾ 32-bit offset permits jump anywhere in memory
Jumps Based on Specific Flags
Jumps Based on Equality
Jumps Based on Unsigned Comparisons
Jumps Based on Signed Comparisons
Block-Structured IF Statements
Assembly language programmers can easily translate logical statements written in C++/Java into assembly language. For example:
mov eax,op1 if( op1 == op2 )
cmp eax,op2 jne L1
mov X,1 jmp L2 L1: mov X,2 L2:
if( op1 op2 ) X = 1;
else
X = 2;
Your Turn . . .
Implement the following pseudocode in
assembly language. All values are unsigned:
cmp ebx,ecx ja next mov eax 5 if( ebx <= ecx )
{ mov eax,5
mov edx,6 next:
eax = 5;
edx = 6;
}
(There are multiple correct solutions to this problem.)
WHILE Loops
while( eax < ebx) eax = eax + 1;
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:
top: cmp eax,ebx ; check loop condition jae next ; false? exit loop
inc eax ; body of loop
jmp top ; repeat the loop
next:
This is a possible implementation:
Your Turn . . .
while( ebx <= val1) {
ebx = ebx + 5;
val1 = val1 - 1
Implement the following loop, using unsigned 32-bit integers:
top: cmp ebx,val1 ; check loop condition ja next ; false? exit loop
add ebx,5 ; body of loop dec val1
jmp top ; repeat the loop
next:
}
DIV Instruction (See 7.4.4)
The DIV (unsigned divide) instruction performs 8- bit, 16-bit, and 32-bit division on unsigned integers
A single operand is supplied (register or memory operand), which is assumed to be the divisor
Instruction formats:
Instruction formats:
DIV r/m8 DIV r/m16 DIV r/m32
Default Operands:
DIV Examples
Divide 8003h by 100h, using 16-bit operands:
mov dx,0 ; clear dividend, high mov ax,8003h ; dividend, low
mov cx,100h ; divisor
div cx ; AX = 0080h, DX = 3
Same division, using 32-bit operands:
mov edx,0 ; clear dividend, high mov eax,8003h ; dividend, low
mov ecx,100h ; divisor
div ecx ; EAX = 00000080h, DX = 3
Your Turn . . .
mov dx,0087h
What will be the hexadecimal values of DX and AX after the following instructions execute?
mov ax,6000h mov bx,100h div bx
DX = 0000h, AX = 8760h
