Ex: Write a piece of code that transfers a block of 256 bytes stored at locations starting at 34000H to locations starting at 36000H. Ans.

Transcription

1 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI Conditional Jump Cond Unsigned Signed = JE : Jump Equal JE : Jump Equal ZF = 1 JZ : Jump Zero JZ : Jump Zero ZF = 1 JNZ : Jump Not Zero JNZ : Jump Not Zero ZF = 0 JNE : Jump Not Equal JNE : Jump Not Equal ZF = 0 > JA : Jump above CF = 0 JG : Jump Greater ZF = 0 and ZF = 0 and SF = OF < JB : Jump Below CF = 1 JL : Jump Less SF OF JAE : Jump Above or Equal JGE : Jump Greater or CF = 0 Equal SF = OF JBE : Jump Below or Equal CF = 1 JLE : Jump Less or Equal SF OF or ZF = 1 or ZF = 1 Carry Zero Parity Sign Overflow JC : Jump if Carry CF=1 JNC: Jump if No Carry CF=0 JZ: Jump if Zero ZF=1 JNZ: Jump if No Zero ZF=0 JPE : Jump if Parity Even PF=1 JPO: Jump if Parity Odd PF=0 JS: Jump if Signed (if negative) SF = 1 JNS: Jump if Not Signed (if positive) SF = 0 JO: Jump if Overflow OF = 1 JNO: Jump if Not Overflow OF = 0 Ex: Write a piece of code that transfers a block of 256 bytes stored at locations starting at 34000H to locations starting at 36000H Ans FF 340FF

2 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI MOV AX, 3000H MOV BX, 0000H NEXT: MOV AL, [BX H] MOV [BX H], AL INC BX JNZ NEXT Ex : Write an ALP that adds an array of integers of size 256 byte stored at locations starting at 54000H with another array starting at address 56000H, store the result at locations starting at 58000H Ans 540FF 560FF 580FF MOV BX, 0000H NEXT: MOV AL, [BX H] + = ADD AL, [BX H] MOV [BX H], AL INC BX JNZ NEXT Ex: Write an ALP to find the absolute subtraction of two arrays of size 1024 bytes starting at 82000H and 84000H respectively Store the results at locations starting at 86000H Ans MOV AX, 8000H XOR SI, SI MOV CX, 400H NEXT: MOV AL, [SI H] CMP AL, [SI H] JB N1 SUB AL, [SI H] JMP N2 N1: MOV BL, [SI H] XCHG AL, BL SUB AL, BL N2: MOV [SI H], AL 34

3 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI INC SI JNZ NEXT Ex: Write an ALP that counts the number of 1 s in a byte stored in 54000H and stores the result in 54001H Ans XOR BL, BL ; clear BL to keep the number of 1s MOV CL, 8 ; rotate total of 8 times MOV SI, 4000H MOV AL, [SI] AGAIN: ROL AL,1 ; rotate it once JNC NEXT ; check for 1 INC BL ; if CF=1 then add one to count NEXT: DEC CL ; go through this 8 times JNZ AGAIN ; if not finished go back INC SI MOV [SI], BL Ex : Write a piece of code to find the number of negative integers in an array of size 1024 byte contain signed numbers stored at addresses starting at 21000H, store the result in a location 51000H Ans MOV AX, 2000H MOV CX, 400H MOV DL, 00H MOV SI, 0000H AGAIN: MOV AL, [SI+1000H] ROL AL, 1 JNC NEXT INC DX NEXT: INC SI JNZ AGAIN MOV [1000H], DX Ex: Write an ALP to find the maximum byte of a block of 256 bytes starting at 53000H Store the result at 56000H 35

4 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI MOV SI, 3000 H MOV CX, 0100 H MOV AH, 00 H NEXT: CMP AH, [SI] JAE PASS MOV AH, [SI] PASS: INC SI JNZ NEXT MOV DI, 6000 MOV [DI], AH Ex : Write a program finds the factorial of 08H Ans MOV BX, 08 MOV AX, 0001 LOOP: MUL BX DEC BX JNZ LOOP HW : Write a piece of code to find the number of odd integers in an array of size 1024 byte stored at addresses starting at 21000H, store the result in location 51000H Write a piece of code that exchanges a block of 256 bytes stored at locations starting at 34000H with another block starting at 36000H By using XCHG instruction By using MOV instruction Write an ALP to find the 2 s complement of a block of 100 bytes starting at 53000H and store the result in Write an ALP to find the minimum value of a byte from a block of 256 bytes starting at 53000H Store the result at 56000H Write an ALP to find the sum of the following series: Sum=

5 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI Branch Program Structure IF THEN ELSE The High level language IF THEN ELSE structure is expressed by the following assembly structure: CMP AX, BX JNE NEQUAL ; Next instruction if (AX) = (BX) NEQUAL: ; Next instruction if (AX) (BX) Loop Program Structures WHILE LOOP A typical WHILE LOOP shown in the following flow chart can be expressed by the following assembly structure: count= final; while (count!=0) { Loop program statements; --count; } // next statement MOV CL, COUNT ; Set loop repeat count AGAIN: JZ NEXT ; Loop is complete if CL=0 (ZF=1) ; First instruction of the loop ; Second instruction of the loop ; last instruction of the loop DEC CL ; Decrement repeat count by 1 JMP AGAIN ; Repeat from AGAIN NEXT: ; First instruction executed after the loop is complete 38

6 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI Loop Instructions There are three loop instructions They can be used in place of certain conditional jump instructions These instructions give the programmer a flexibility in writing programs in a simpler manner The different loop instructions and the operations they perform are listed below: CX = CX - 1 CX 0? Yes NEXT No Ex: Write a piece of code that transfers a block of 256 bytes stored at locations starting at 34000H to locations starting at 36000H Ans MOV AX, 3000H MOV BX, 0000H NEXT: MOV AL, [BX H] MOV [BX H], AL INC BX LOOP NEXT 30

7 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI Ex: Write an ALP to find the maximum byte from a block of 256 bytes starting at 53000H Ans LEA SI, [3000H] MOV CX, 0100H MOV AH, 00H AGAIN: CMP AH, [SI] JAE NEXT MOV AH, [SI] NEXT: INC SI LOOPNE AGAIN MOV [SI], AH Ex : Write an ALP that counts the number of letters M in a string of size 256 bytes starts at 5F600H LEA SI, [F600H] MOV DL, 00H MOV AL, 4DH ; 4D is the ASCII code of the letter M AGAIN: CMP AL, [SI] JNE NEXT INC DL NEXT: INC SI LOOP AGAIN Ex: Write an ALP to find the average of an array of size 256 bytes of unsigned integers starting at 6A000H MOV AX, 6000H XOR BX, BX XOR AX, AX AGAIN: ADD AL, [BX+2000H] ADC AH, 00H NEXT: INC BX LOOP AGAIN XOR DX, DX DIV BX HW: Repeat the previous example if the array has signed integers 39

8 INSTRUCTOR: ABDULMUTTALIB A H ALDOURI Stack Instructions The stack in the 8086/8088 µps, like that in many microprocessors, is a region of memory that can store information temporarily مؤقتت) (بصىرة during the execution of the program It is called a stack, because you "stack" things on it The stack is a Last-In-First-Out, (LIFO) structure so the last thing stored in (يستزجع) the stack is the first thing retrieved The philosophy is that you retrieve (pop) data in the opposite order of storing (push) it In the 8086/8088µP, the stack pointer is SS:SP (physical address), which is a 16 bit pointer into a 20 bit address space In a POP operation, the data is retrieved (POPed) from that address (SS : SP) The SP register is incremented by 2 (Low byte of the operand) SS : SP (High byte of the operand) SS : (SP) + 1 (SP) (SP) + 2 In a PUSH operation, the data of the source is stored (pushed) at address (SS : SP) The SP register is decremented 2 SS: (SP) 1 (High byte of the operand) SS: (SP) 2 (Low byte of the operand) (SP) (SP) 2 POPF retrieves a word from the stack and places it into the flags register and increments the stack pointer SP by 2 (Low Byte of Flags Reg) SS : SP (High Byte of Flags Reg) SS : (SP) + 1 (SP) (SP) + 2 PUSHF pushes the contents of the flags register onto the stack at address (SS : SP) and decrements the stack pointer SP by 2 SS : (SP) 1 (High Byte of Flags Reg) SS : (SP) 2 (Low Byte of Flags Reg) (SP) (SP) 2 48