CSC 204 - Assembly Language - Chapter 04 - Fall 2024-2025 PDF

Assembly Language Programming Chapter 4 Rabih Tarraf Data Transfers, Addressing,...

Assembly Language Programming Chapter 4 Rabih Tarraf Data Transfers, Addressing, and arithmetic Chapter Overview 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 rabih@tarraf 2 1  Chapter Overview Data Transfer Instructions Addition and Subtraction Data-Related Operators and Directives Indirect Addressing JMP and LOOP Instructions 64-Bit Programming rabih@tarraf 3 Data Transfer Instructions Operand Types Instruction Operand Notation Direct Memory Operands MOV Instruction Zero & Sign Extension XCHG Instruction Direct-Offset Instructions rabih@tarraf 4 2  Operand Types Instructions can have zero, one, two, or three operands. Here, we omit the label and comment fields for clarity: There are three basic types of operands: Immediate operand — uses a numeric or character literal expression [a constant integer (8, 16, or 32 bits)] Register operand — uses a named CPU register [the name of a register (eax, ebx, rax,… etc)] Memory operand — references a memory location [reference to a location in memory] rabih@tarraf 5 Instruction Operand Notation rabih@tarraf 6 3  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 Offset Data var1 BYTE 10h 10400h 10h Var1 1 Byte.code mov al,var1 ; AL = 10h mov al,[var1] ; AL = 10h C# code: Byte var1 = 16 Assembly Code: mov al, var1 alternate format Machine Code: A0 00010400 Opcode Memory @ (32 bits) rabih@tarraf 7 MOV Instruction Move from source to destination. Syntax: MOV destination, source ; data transfer instruction: copies data from a source operand to a destination operand Both operands must be the same size ( reg , imm is exception). 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 reg,mem mov ax,wVal ; mov reg,mem mov count,al ; mov mem,reg mov al,wVal ; error mov ax,count ; error mov eax,count ; error rabih@tarraf 8 4  Overlapping Values The following code example shows how the same 32-bit register can be modified using differently sized data. When oneWord is moved to AX, it overwrites the existing value of AL. When oneDword is moved to EAX, it overwrites AX. Finally, when 0 is moved to AX, it overwrites the lower half of EAX. rabih@tarraf 9 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 immediate move to DS not permitted mov esi,wVal size mismatch mov eip,dVal EIP cannot be the destination mov 25,bVal immediate value cannot be destination mov bVal2,bVal memory-to-memory move not permitted rabih@tarraf 10 5  Copying Smaller Values to Larger Ones? What happens if we try the same approach with a signed integer equal to -16? rabih@tarraf 11 Zero Extension The MOVZX instruction (move with zero-extend) copies the contents of a source operand into a destination operand and zero-extends the value to 16 or 32 bits. This instruction is only used with unsigned integers. There are three variants: mov bl,10001111b movzx ax,bl ; zero-extension The destination must be a register. rabih@tarraf 12 6  Example rabih@tarraf 13 Sign Extension The MOVSX instruction (move with sign-extend) copies the contents of a source operand into a destination operand and sign-extends the value to 16 or 32 bits. This instruction is only used with signed integers. mov bl,10001111b movsx ax,bl ; sign extension The destination must be a register. rabih@tarraf 14 7  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 ; 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 rabih@tarraf 17 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 ; AL = 20h mov al,[arrayB+1] ; alternative notation Q: Why doesn't arrayB+1 produce 11h? rabih@tarraf 18 8  Direct-Offset Operands (cont) 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? rabih@tarraf 19 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 rabih@tarraf 20 9  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? rabih@tarraf 21 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 mov bl,[myBytes+2] add ax,bx ; AX = sum Yes: Move zero to BX before the MOVZX instruction. rabih@tarraf 22 10  Section Review (S01) 1. What are the three basic types of operands? 2. (True/False): The destination operand of a MOV instruction cannot be a segment register. 3. (True/False): In a MOV instruction, the second operand is known as the destination operand. 4. (True/False): The EIP register cannot be the destination operand of a MOV instruction. 5. In the operand notation used by Intel, what does reg/mem32 indicate? 6. In the operand notation used by Intel, what does imm16 indicate? rabih@tarraf 23 What's Next Data Transfer Instructions Addition and Subtraction Data-Related Operators and Directives Indirect Addressing JMP and LOOP Instructions 64-Bit Programming rabih@tarraf 24 11  Addition and Subtraction INC and DEC Instructions ADD and SUB Instructions NEG Instruction Implementing Arithmetic Expressions Flags Affected by Arithmetic Zero Sign Carry Overflow rabih@tarraf 25 INC and DEC Instructions Add 1, subtract 1 from destination operand operand may be register or memory INC destination Logic: destination  destination + 1 DEC destination Logic: destination  destination – 1 rabih@tarraf 26 12  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 rabih@tarraf 27 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 = FFh mov ah,[myByte+1] ; AH = 00h dec ah ; AH = FFh inc al ; AL = 00h dec ax ; AX = FEFF rabih@tarraf 28 13  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 rabih@tarraf 29 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 rabih@tarraf 30 14  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 ; 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? rabih@tarraf 31 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 ; CF = 1, OF = 0 neg [valB + 1] ; CF = 0, OF = 0 neg valC ; CF = 1, OF = 1 rabih@tarraf 32 15  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 ; EAX = -26 mov ebx,Yval sub ebx,Zval ; EBX = -10 add eax,ebx mov Rval,eax ; -36 rabih@tarraf 33 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 rabih@tarraf 34 16  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. rabih@tarraf 35 Concept Map CPU part of executes executes ALU conditional jumps arithmetic & bitwise used by operations attached to provide affect status flags branching logic You can use diagrams such as these to express the relationships between assembly language concepts. rabih@tarraf 36 17  Zero Flag (ZF) The Zero flag is set when the result of an operation produces zero in the destination operand. mov cx,1 sub cx,1 ; CX = 0, ZF = 1 mov ax,0FFFFh inc ax ; AX = 0, ZF = 1 inc ax ; AX = 1, ZF = 0 Remember... A flag is set when it equals 1. A flag is clear when it equals 0. rabih@tarraf 37 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 ; CX = -1, SF = 1 add cx,2 ; CX = 1, SF = 0 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 rabih@tarraf 38 18  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 rabih@tarraf 39 Overflow and Carry Flags A Hardware Viewpoint How the ADD instruction affects OF and CF: CF = (carry out of the MSB) OF = CF XOR MSB How the SUB instruction affects OF and CF: CF = INVERT (carry out of the MSB) negate the source and add it to the destination OF = CF XOR MSB MSB = Most Significant Bit (high-order bit) XOR = eXclusive-OR operation NEG = Negate (same as SUB 0,operand ) rabih@tarraf 40 19  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 rabih@tarraf 41 Your turn... For each of the following marked entries, show the values of the destination operand and the Sign, Zero, and Carry flags: mov ax,00FFh add ax,1 ; AX=0100h SF=0 ZF=0 CF=0 sub ax,1 ; AX=00FFh SF=0 ZF=0 CF=0 add al,1 ; AL=00h SF=0 ZF=1 CF=1 mov bh,6Ch add bh,95h ; BH=01h SF=0 ZF=0 CF=1 mov al,2 sub al,3 ; AL=FFh SF=1 ZF=0 CF=1 rabih@tarraf 42 20  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 ; OF = 1, AL = 80h add al,1 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. rabih@tarraf 43 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 rabih@tarraf 44 21  Your turn... What will be the values of the given flags after each operation? mov al,-128 neg al ; CF = 1 OF = 1 mov ax,8000h add ax,2 ; CF = 0 OF = 0 mov ax,0 sub ax,2 ; CF = 1 OF = 0 mov al,-5 sub al,+125 ; OF = 1 rabih@tarraf 45 Section Review (S02) rabih@tarraf 46 22  What's Next Data Transfer Instructions Addition and Subtraction Data-Related Operators and Directives Indirect Addressing JMP and LOOP Instructions 64-Bit Programming rabih@tarraf 47 Data-Related Operators and Directives OFFSET Operator PTR Operator TYPE Operator LENGTHOF Operator SIZEOF Operator LABEL Directive rabih@tarraf 48 23  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 use only a single segment (flat memory model). rabih@tarraf 49 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 bVal ; ESI = 00404000 mov esi,OFFSET wVal ; ESI = 00404001 mov esi,OFFSET dVal ; ESI = 00404003 mov esi,OFFSET dVal2 ; ESI = 00404007 rabih@tarraf 50 24  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: ; Assembly language: char array;.data char * p = array; array BYTE 1000 DUP(?).code mov esi,OFFSET array rabih@tarraf 51 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 ; error – why? mov ax,WORD PTR myDouble ; loads 5678h mov WORD PTR myDouble,4321h ; saves 4321h Little endian order is used when storing data in memory (see Section 3.4.9). rabih@tarraf 52 25  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: doubleword word byte offset 12345678 5678 78 0000 myDouble When integers are loaded from 56 0001 myDouble memory +1 into registers, the bytes are automatically re-reversed into their 1234 34 0002 myDouble correct +2 positions. 12 0003 myDouble + 3 rabih@tarraf 53 PTR Operator Examples.data myDouble DWORD 12345678h doubleword word byte offset 12345678 5678 78 0000 myDouble 56 0001 myDouble + 1 1234 34 0002 myDouble + 2 12 0003 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 rabih@tarraf 54 26  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] ; AX = 3412h mov ax,WORD PTR [myBytes+2] ; AX = 7856h mov eax,DWORD PTR myBytes ; EAX = 78563412h rabih@tarraf 55 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] ; a. 0502h mov bl,BYTE PTR varD ; b. 78h mov bl,BYTE PTR [varW+2] ; c. 02h mov ax,WORD PTR [varD+2] ; d. 1234h mov eax,DWORD PTR varW ; e. 12026543h rabih@tarraf 56 27  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 rabih@tarraf 57 LENGTHOF Operator The LENGTHOF operator counts the number of elements in a single data declaration..data LENGTHOF byte1 BYTE 10,20,30 ; 3 array1 WORD 30 DUP(?),0,0 ; 32 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 rabih@tarraf 58 28  SIZEOF Operator The SIZEOF operator returns a value that is equivalent to multiplying LENGTHOF by TYPE..data SIZEOF byte1 BYTE 10,20,30 ; 3 array1 WORD 30 DUP(?),0,0 ; 64 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 rabih@tarraf 59 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 ; 6 mov ebx,SIZEOF array ; 12 rabih@tarraf 60 29  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 ; 2 mov ebx,SIZEOF array ; 4 rabih@tarraf 61 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 rabih@tarraf 62 30  Section Review (S03) 1. (True/False): The OFFSET operator always returns a 16-bit value. 2. (True/False): The PTR operator returns the 32-bit address of a variable. 3. (True/False): The TYPE operator returns a value of 4 for doubleword operands. 4. (True/False): The LENGTHOF operator returns the number of bytes in an operand. 5. (True/False): The SIZEOF operator returns the number of bytes in an operand. rabih@tarraf 63 What's Next Data Transfer Instructions Addition and Subtraction Data-Related Operators and Directives Indirect Addressing JMP and LOOP Instructions 64-Bit Programming rabih@tarraf 64 31  Indirect Addressing Indirect Operands Array Sum Example Indexed Operands Pointers rabih@tarraf 65 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] ; dereference ESI (AL = 10h) inc esi mov al,[esi] ; AL = 20h inc esi mov al,[esi] ; AL = 30h rabih@tarraf 66 32  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] ; error: ambiguous inc WORD PTR [esi] ; ok Should PTR be used here? yes, because [esi] could point to a byte, word, or doubleword add [esi],20 rabih@tarraf 67 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 WORD 1000h,2000h,3000h.code 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. rabih@tarraf 68 33  Indexed Operands An indexed operand adds a constant to a register to generate an effective address. There are two notational forms: [label + reg] label[reg].data arrayW WORD 1000h,2000h,3000h.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. rabih@tarraf 69 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] ; 04 mov bx,arrayW[esi*TYPE arrayW] ; 0004 mov edx,arrayD[esi*TYPE arrayD] ; 00000004 rabih@tarraf 70 34  Pointers 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: ptrW DWORD OFFSET arrayW rabih@tarraf 71 Section Review (S04) 1. (True/False): Any 32-bit general-purpose register can be used as an indirect operand. 2. (True/False): The EBX register is usually reserved for addressing the stack. 3. (True/False): The following instruction is invalid: inc [esi] 4. (True/False): The following is an indexed operand: array[esi] rabih@tarraf 72 35 Section Review (S04) rabih@tarraf 73 Section Review (S04) rabih@tarraf 74 36  What's Next Data Transfer Instructions Addition and Subtraction Data-Related Operators and Directives Indirect Addressing JMP and LOOP Instructions 64-Bit Programming rabih@tarraf 75 JMP and LOOP Instructions JMP Instruction LOOP Instruction LOOP Example Summing an Integer Array Copying a String rabih@tarraf 76 37  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 5.5.2.3 for details). rabih@tarraf 77 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. rabih@tarraf 78 38  LOOP Example The following loop calculates the sum of the integers 5 + 4 + 3 +2 + 1: offset machine code source code 00000000 66 B8 0000 mov ax,0 00000004 B9 00000005 mov ecx,5 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 rabih@tarraf 79 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 rabih@tarraf 80 39  Your turn... mov ax,6 mov ecx,4 What will be the final value of AX? L1: inc ax 10 loop L1 mov ecx,0 How many times will the loop X2: execute? inc ax 4,294,967,296 loop X2 rabih@tarraf 81 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 ; 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 rabih@tarraf 82 40  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 ; 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 rabih@tarraf 83 Your turn... What changes would you make to the program on the previous slide if you were summing a doubleword array? rabih@tarraf 84 41  Copying a String The following code copies a string from source to target:.data source BYTE "This is the source string",0 good use of target BYTE SIZEOF source DUP(0) SIZEOF.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 rabih@tarraf 85 Your turn... Rewrite the program shown in the previous slide, using indirect addressing rather than indexed addressing. rabih@tarraf 86 42 Section Review (S05) rabih@tarraf 87 Section Review (S05) rabih@tarraf 88 43  What's Next Data Transfer Instructions Addition and Subtraction Data-Related Operators and Directives Indirect Addressing JMP and LOOP Instructions 64-Bit Programming rabih@tarraf 89 64-Bit Programming MOV instruction in 64-bit mode accepts operands of 8, 16, 32, or 64 bits When you move a 8, 16, or 32-bit constant to a 64-bit register, the upper bits of the destination are cleared. When you move a memory operand into a 64-bit register, the results vary: 32-bit move clears high bits in destination 8-bit or 16-bit move does not affect high bits in destination rabih@tarraf 90 44  More 64-Bit Programming MOVSXD sign extends a 32-bit value into a 64-bit destination register The OFFSET operator generates a 64-bit address LOOP uses the 64-bit RCX register as a counter RSI and RDI are the most common 64-bit index registers for accessing arrays. rabih@tarraf 91 Other 64-Bit Notes ADD and SUB affect the flags in the same way as in 32-bit mode You can use scale factors with indexed operands. rabih@tarraf 92 45  Section Review (S06) rabih@tarraf 93 Summary 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 rabih@tarraf 94 46 Ch 3 rabih@tarraf 96 Ch 4 rabih@tarraf 97 47 Ch 4 rabih@tarraf 98 Ch 6 rabih@tarraf 99 48  rabih@tarraf 100 Chapter rabih@tarraf 49 Good References rabih@tarraf 50