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Using programming knowledge, it is possible to use programming logic, we can differentiate and find the terms
.data
sourceword:
.word 0xAB
wordmask:
.word 0xCF
operand:
.long 0xAA
.text
.globl _start
_start:
movw sourceword, %ax # ax=0x00AB ( move the word value of sourceword =0x00AB to ax register.)
movw %ax, %bx # bx=0x00AB ( copy the value of ax to bx register.)
xorw %cx, %cx # cx=0000 ( cx=cx XOR cx, that is clear cx register.)
andw wordmask, %ax # ax=0x008B ( ax=ax AND wordmask= 0x00AB AND 0x00CF= 0x008B)
orw wordmask, %bx # bx=0x00EF (bx=bx OR wordmask= 0x00AB OR 0x00CF= 0x00EF.)
xorw wordmask, %cx # cx=0x00CF (cx=cx XOR wordmask= 0x0000 XOR 0x00CF=0x00CF.)
movl operand, %eax # eax=0x000000AA. ( copy the 32-bit value of operand=0x000000AA to eax.)
movl %eax, %ebx # ebx=0x000000AA.( copy the value of eax to ebx.)
movl %eax, %ecx # ecx=0x000000AA. ( copy the value of eax to ecx.)
shll $3,%eax # eax=0x00000550. ( logical shift left of eax=0x000000AA by 3 bits.)
# 0x000000AA=00000000000000000000000010101010
# =>00000000000000000000010101010000=0x00000550.
rorl $4,%ebx # ebx=0xA000000A. (Rotate right ebx=0x000000AA by 4 bits.)
# 0x000000AA=00000000000000000000000010101010
3 =>10100000000000000000000000001010= 0xA000000A.
sarl %ecx # here the count value to shift the bits not mentioned.
# It is the arithmetic shift right instruction. the shifted left bits filled with MSB bit.
See more about logic program at brainly.com/question/14970713
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