大学生考试网 让学习变简单
当前位置:首页 >> 理学 >>



Chapter 4: Data Transfers, Addressing, and Arithmetic

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:
– Immediate – a constant integer (8, 16, or 32 bits)

value is encoded within the instruction register name is converted to a number and encoded within the instruction

– Register – the name of a register

– 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 mov al,[var1]

; AL = 10h ; AL = 10h

alternate format


MOV Instruction
? Move from source to destination. Syntax: MOV destination,source ? No more than one memory operand permitted ? CS, EIP, and IP cannot be the destination ? No immediate to segment moves
.data count BYTE 100 wVal WORD 2 .code mov bl,count mov ax,wVal mov count,al mov al,wVal mov ax,count mov eax,count ; error ; error ; error


Your turn . . .
Explain why each of the following MOV statements are invalid:
.data bVal BYTE 100 bVal2 BYTE ? 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


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.
0 10001111 Source




mov bl,10001111b movzx ax,bl ; zero-extension

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.
10001111 Source




mov bl,10001111b movsx ax,bl ; sign extension

The destination must be a register.


XCHG Instruction
XCHG exchanges the values of two operands. At least one operand must be a register. No immediate operands are permitted.
.data var1 WORD 1000h var2 WORD 2000h .code xchg ax,bx xchg ah,al xchg var1,bx xchg eax,ebx xchg var1,var2

; ; ; ;

exchange exchange exchange exchange

16-bit regs 8-bit regs mem, reg 32-bit regs

; error: two memory operands

Direct-Offset Operands
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.
.data arrayB BYTE 10h,20h,30h,40h .code mov al,arrayB+1 mov al,[arrayB+1]

; AL = 20h ; alternative notation

Q: Why doesn't arrayB+1 produce 11h?

Direct-Offset Operands
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.
.data arrayW WORD 1000h,2000h,3000h arrayD DWORD 1,2,3,4 .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.
mov eax,arrayD xchg eax,[arrayD+4]

? Step 2: Exchange EAX with the third array value and copy the value in EAX to the first array position.
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]

? 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 mov add mov add ax,myBytes bl,[myBytes+1] ax,bx bl,[myBytes+2] 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
– Zero – Sign – Carry – Overflow


INC and DEC Instructions

Add 1, subtract 1 from destination operand
– operand may be register or memory


INC destination

Logic: destination ? destination + 1 Logic: destination ? destination – 1


DEC destination


INC and DEC Examples
.data myWord WORD 1000h myDword DWORD 10000000h .code inc myWord dec myWord inc myDword mov inc mov inc ax,00FFh ax ax,00FFh al

; 1001h ; 1000h ; 10000001h

; AX = 0100h ; AX = 0000h


Your turn...
Show the value of the destination operand after each of the following instructions executes:
.data myByte .code mov mov dec inc dec BYTE 0FFh, 0 al,myByte ah,[myByte+1] ah al ax ; ; ; ; ; AL AH AH AL 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 instruction


ADD and SUB Examples
.data var1 DWORD 10000h var2 DWORD 20000h .code mov eax,var1 add eax,var2 add ax,0FFFFh add eax,1 sub ax,1

; ; ; ; ; ;

---EAX--00010000h 00030000h 0003FFFFh 00040000h 0004FFFFh


NEG (negate) Instruction
Reverses the sign of an operand. Operand can be a register or memory operand.
.data valB BYTE -1 valW WORD +32767 .code mov al,valB neg al neg valW

; AL = -1 ; AL = +1 ; valW = -32767

Suppose AX contains –32,768 and we apply NEG to it. Will the result be valid?


NEG Instruction and the Flags
The processor implements NEG using the following internal operation:
SUB 0,operand

Any nonzero operand causes the Carry flag to be set.
.data valB BYTE 1,0 valC SBYTE -128 .code neg valB neg [valB + 1] neg valC

; CF = 1, OF = 0 ; CF = 0, OF = 0 ; CF = 1, OF = 1

Implementing Arithmetic Expressions
HLL compilers translate mathematical expressions into assembly language. You can do it also. For example:
Rval = -Xval + (Yval – Zval)
Rval DWORD ? Xval DWORD 26 Yval DWORD 30 Zval DWORD 40 .code mov eax,Xval neg eax mov ebx,Yval sub ebx,Zval add eax,ebx mov Rval,eax

; EAX = -26 ; EBX = -10 ; -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 neg add mov sub mov ebx,Yval ebx ebx,Zval eax,Xval eax,ebx Rval,eax

Flags Affected by Arithmetic



The ALU has a number of status flags that reflect the outcome of arithmetic (and bitwise) operations – based on the contents of the destination operand Essential flags: – Zero flag – set when destination equals zero – Sign flag – set when destination is negative – Carry flag – set when unsigned value is out of range – Overflow flag – set when signed value is out of range The MOV instruction never affects the flags.

Concept Map
part of executes executes


conditional jumps
arithmetic & bitwise operations
attached to used by provide

status flags branching logic

You can use diagrams such as these to express the relationships between assembly language concepts.


Zero Flag (ZF)
The Zero flag is set when the result of an operation produces zero in the destination operand.
mov sub mov inc inc cx,1 cx,1 ax,0FFFFh ax ax ; CX = 0, ZF = 1 ; AX = 0, ZF = 1 ; AX = 1, ZF = 0

Remember... ? A flag is set when it equals 1. ? A flag is clear when it equals 0.

Sign Flag (SF)
The Sign flag is set when the destination operand is negative. The flag is clear when the destination is positive.
mov cx,0 sub cx,1 add cx,2

; CX = -1, SF = 1 ; CX = 1, SF = 0

The sign flag is a copy of the destination's highest bit:
mov al,0 sub al,1 add al,2 ; AL = 11111111b, SF = 1 ; 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 using the correct data type with each instruction


Overflow and Carry Flags A Hardware Viewpoint

How the ADD instruction modifies OF and CF: – OF = (carry out of the MSB) XOR (carry into the MSB) – CF = (carry out of the MSB) How the SUB instruction modifies OF and CF: – NEG the source and ADD it to the destination – OF = (carry out of the MSB) XOR (carry into the MSB) – 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 ; Try to go below zero: mov al,0 sub al,1 ; CF = 1, AL = 00

; CF = 1, AL = FF

Your turn . . .
For each of the following marked entries, show the values of the destination operand and the Sign, Zero, and Carry flags:
mov add sub add mov add ax,00FFh ax,1 ax,1 al,1 bh,6Ch bh,95h ; AX= 0100h SF= 0 ZF= 0 CF= 0 ; AX= 00FFh SF= 0 ZF= 0 CF= 0 ; AL= 00h SF= 0 ZF= 1 CF= 1 ; BH= 01h SF= 0 ZF= 0 CF= 1

mov al,2 sub al,3

; AL= FFh

SF= 1 ZF= 0 CF= 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. 36

A Rule of Thumb

When adding two integers, remember that the Overflow flag is only set when . . . – Two positive operands are added and their sum is negative – Two negative operands are added and their sum is positive
What will be the values of the Overflow flag? mov al,80h add al,92h ; OF = 1 mov al,-2 add al,+127 ; OF = 0

Your turn . . .
What will be the values of the given flags after each operation?
mov al,-128 neg al mov ax,8000h add ax,2 mov ax,0 sub ax,2

; CF = 1

OF = 1

; CF = 0
; CF = 1 ; OF = 1

OF = 0
OF = 0

mov al,-5 sub al,+125


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
– Protected mode: 32 bits – Real mode: 16 bits
offset data segment: myByte

The Protected-mode programs we write only have a single segment (we use the flat memory model).


OFFSET Examples
Let's assume that the data segment begins at 00404000h:

.data bVal BYTE ? wVal WORD ? dVal DWORD ? dVal2 DWORD ? .code mov esi,OFFSET mov esi,OFFSET mov esi,OFFSET mov esi,OFFSET

bVal wVal dVal dVal2

; ; ; ;


= = = =

00404000 00404001 00404003 00404007

Relating to C/C++
The value returned by OFFSET is a pointer. Compare the following code written for both C++ and assembly language:
; C++ version: char array[1000]; char * p = array;

.data array BYTE 1000 DUP(?) .code mov esi,OFFSET array

; ESI is p

PTR Operator
Overrides the default type of a label (variable). Provides the flexibility to access part of a variable.
.data myDouble DWORD 12345678h .code mov ax,myDouble
mov ax,WORD PTR myDouble mov WORD PTR myDouble,4321h

; error – why? ; loads 5678h ; 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:
byte offset



78 5678

78 56


34 12

When integers are loaded from memory into registers, the myDouble + 1 0001 bytes are automatically re0002 myDouble + 2 reversed into their correct positions. 0003 myDouble + 3


PTR Operator Examples
.data myDouble DWORD 12345678h





12345678 5678

78 56

0000 0001 0002 0003

myDouble myDouble + 1 myDouble + 2 myDouble + 3


34 12

mov mov mov mov mov


PTR myDouble PTR [myDouble+1] PTR [myDouble+2] PTR myDouble PTR [myDouble+2]

; ; ; ; ;


= = = = =

78h 56h 34h 5678h 1234h


PTR Operator (cont)
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.

.data myBytes BYTE 12h,34h,56h,78h .code mov ax,WORD PTR [myBytes] mov ax,WORD PTR [myBytes+2] mov eax,DWORD PTR myBytes

; AX = 3412h ; AX = 7856h ; EAX = 78563412h

Your turn . . .
Write down the value of each destination operand:
.data varB BYTE 65h,31h,02h,05h varW WORD 6543h,1202h varD DWORD 12345678h .code mov ax,WORD PTR [varB+2] mov bl,BYTE PTR varD mov cl,BYTE PTR [varW+2] mov dx,WORD PTR [varD+2] mov eax,DWORD PTR varW

; ; ; ; ;

ax = bl = cl = dx = eax =

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 mov eax,TYPE mov eax,TYPE mov eax,TYPE

var1 var2 var3 var4

; ; ; ;

1 2 4 8


The LENGTHOF operator counts the number of elements in a single data declaration.
.data byte1 BYTE 10,20,30 array1 WORD 30 DUP(?),0,0 array2 WORD 5 DUP(3 DUP(?)) array3 DWORD 1,2,3,4 digitStr BYTE "12345678",0 .code mov ecx,LENGTHOF array1 LENGTHOF ; 3 ; 32 ; 15 ; 4 ; 9

; 32


SIZEOF Operator
The SIZEOF operator returns a value that is equivalent to multiplying LENGTHOF by TYPE.
.data byte1 BYTE 10,20,30 array1 WORD 30 DUP(?),0,0 array2 WORD 5 DUP(3 DUP(?)) array3 DWORD 1,2,3,4 digitStr BYTE "12345678",0 SIZEOF ; 3 ; 64 ; 30 ; 16 ; 9

.code mov ecx,SIZEOF array1

; 64


Spanning Multiple Lines
(1 of 2)
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 mov ebx,SIZEOF array

; 6 ; 12

Spanning Multiple Lines
(2 of 2)
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 mov ebx,SIZEOF array

; 2 ; 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 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 (1 of 2)
An indirect operand holds the address of a variable, usually an array or string. It can be dereferenced (just like a pointer).
.data val1 BYTE 10h,20h,30h .code mov esi,OFFSET val1 mov al,[esi] inc esi mov al,[esi] inc esi mov al,[esi]

; dereference ESI (AL = 10h)

; AL = 20h

; AL = 30h

Indirect Operands (2 of 2)
Use PTR to clarify the size attribute of a memory operand.
.data myCount WORD 0 .code mov esi,OFFSET myCount inc [esi] inc WORD PTR [esi]

; error: ambiguous ; ok

Should PTR be used here?
add [esi],20

yes, because [esi] could point to a byte, word, or doubleword

Array Sum Example
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.
.data arrayW .code mov mov add add add add WORD 1000h,2000h,3000h esi,OFFSET arrayW ax,[esi] esi,2 ax,[esi] esi,2 ax,[esi]

; or: add esi,TYPE arrayW

; AX = sum of the array

ToDo: Modify this example for an array of doublewords.


Indexed Operands
An indexed operand adds a constant to a register to generate an effective address. There are two notational forms:
[label + reg] .data arrayW WORD 1000h,2000h,3000h .code mov esi,0 mov ax,[arrayW + esi] mov ax,arrayW[esi] add esi,2 add ax,[arrayW + esi] etc. label[reg]

; AX = 1000h ; alternate format

ToDo: Modify this example for an array of doublewords.


Index Scaling*
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:
.data arrayB BYTE 0,1,2,3,4,5 arrayW WORD 0,1,2,3,4,5 arrayD DWORD 0,1,2,3,4,5 .code mov esi,4 mov al,arrayB[esi*TYPE arrayB] mov bx,arrayW[esi*TYPE arrayW] mov edx,arrayD[esi*TYPE arrayD]

; 04 ; 0004 ; 00000004

* Not in the book

You can declare a pointer variable that contains the offset of another variable.
.data arrayW WORD 1000h,2000h,3000h ptrW DWORD arrayW .code mov esi,ptrW mov ax,[esi] ; AX = 1000h

Alternate format:

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
? Example:
top: . . jmp top

A jump outside the current procedure must be to a special type of label called a global label (see Section 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
The following loop calculates the sum of the integers 5 + 4 + 3 +2 + 1:
offset 00000000
00000004 00000009 0000000C 0000000E

machine code 66 B8 0000
B9 00000005 66 03 C1 E2 FB

source code mov ax,0
mov ecx,5 L1: add ax,cx loop L1

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 . . .
mov ax,6 mov ecx,4

What will be the final value of AX?

inc ax loop L1

How many times will the loop execute? 4,294,967,296

mov ecx,0 X2: inc ax loop X2


Nested Loop
If you need to code a loop within a loop, you must save the outer loop counter's ECX value. In the following example, the outer loop executes 100 times, and the inner loop 20 times.
.data count DWORD ? .code mov ecx,100 L1: mov count,ecx mov ecx,20 L2: . . loop L2 mov ecx,count loop L1

; set outer loop count

; save outer loop count ; set inner loop count

; repeat the inner loop ; restore outer loop count ; repeat the outer loop


Summing an Integer Array
The following code calculates the sum of an array of 16-bit integers.

.data intarray WORD 100h,200h,300h,400h .code mov edi,OFFSET intarray mov ecx,LENGTHOF intarray mov ax,0 L1: add ax,[edi] add edi,TYPE intarray loop L1

; address of intarray ; loop counter ; zero the accumulator ; add an integer ; point to next integer ; 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
The following code copies a string from source to target:
.data source target .code mov mov L1: mov mov inc loop


"This is the source string",0 SIZEOF source DUP(0)

good use of SIZEOF

esi,0 ecx,SIZEOF source al,source[esi] target[esi],al esi L1

; index register ; loop counter ; ; ; ; get char from source store it in the target move to next character repeat for entire string


Your turn . . .
Rewrite the program shown in the previous slide, using indirect addressing rather than indexed addressing.







Data Transfer – MOV – data transfer from source to destination – MOVSX, MOVZX, XCHG Operand types – direct, direct-offset, indirect, indexed Arithmetic – INC, DEC, ADD, SUB, NEG – Sign, Carry, Zero, Overflow flags Operators – OFFSET, PTR, TYPE, LENGTHOF, SIZEOF, TYPEDEF JMP and LOOP – branching instructions

The End


网站首页 | 网站地图 | 学霸百科 | 新词新语
All rights reserved Powered by 大学生考试网 9299.net