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AAM: ASCII Adjust for Multiplication after execution. This instruction converts the product available in the AL into unpacked BCD format. This follows a multiplication instruction. The lower byte of result (unpacked) remains in the AL and the higher byte of result remains in AH. The instance given below describes execution of the instruction. Suppose, a product is available in AL, say AL = 5D. AAM instruction will make unpacked BCD result in the AX. DH is higher than value 9, so add 6 (0110) to it D + 6 = 13H. LSD of 13H is the lower unpacked byte for the result. Increment AH by 1, 5+1 = 6 will be the upper unpacked byte of the result. So after the execution, AH = 06 and AL = 03.
Assume that the registers are initialized to EAX=12345h,EBX =9528h ECX=1275h,EDX=3001h sub AH,AH sub DH,DH mov DL,AL mov CL,3 shl DX,CL shl AX,1 add DX,AX
how i can write a program to divide 2 numbers
Compute the Fibonacci sequence - assembly program: Problem: Fibonacci In this problem you will write a program that will compute the first 20 numbers in the Fibonacci sequ
NEG: Negate:- The negate instruction forms the 2's complement of the particular destination in the instruction. For obtaining 2's complement, it subtracts the contents of destinat
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#include"lcd.asm" ; assembly file is included for displaying lcd characters Main: PORTA EQU 0xF80 ; PORTS PORTB EQU 0xF81 PORTC EQU 0xF82 PORTD EQU 0xF83 R
a- Trace the following program fragment and find out the content of ax after the the execution of the program. X db 5,7 -3,-9,4,-7,9 Mov
SHORT : The SHORT operator denoted to the assembler that only one byte is needed to code the displacement for a jump (for example displacement is within -128 to +127 bytes fr
Architecture Of 8088 The register set of 8088 is accurately the same as in to 8086. The architecture of 8088 is also same to 8086 except for 2 changes; a) 8088 has 4-byte instr
DMA controller : Steps include in transferring a block of data from I/O devices (for example a disk) to memory: 1. CPU sends a signal to initiate disk transfe
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