How many MB are there in 1000 KB?
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Using programming knowledge, it is possible to use programming logic, we can differentiate and find the terms
<h3>Writing the code and understanding the programming logic:</h3><em>.data</em>
<em>sourceword:</em>
<em>.word 0xAB</em>
<em>wordmask:</em>
<em>.word 0xCF</em>
<em>operand:</em>
<em>.long 0xAA</em>
<em>.text</em>
<em>.globl _start</em>
<em>_start:</em>
<em>movw sourceword, %ax # ax=0x00AB ( move the word value of sourceword =0x00AB to ax register.)</em>
<em>movw %ax, %bx # bx=0x00AB ( copy the value of ax to bx register.)</em>
<em>xorw %cx, %cx # cx=0000 ( cx=cx XOR cx, that is clear cx register.)</em>
<em />
<em>andw wordmask, %ax # ax=0x008B ( ax=ax AND wordmask= 0x00AB AND 0x00CF= 0x008B)</em>
<em>orw wordmask, %bx # bx=0x00EF (bx=bx OR wordmask= 0x00AB OR 0x00CF= 0x00EF.)</em>
<em>xorw wordmask, %cx # cx=0x00CF (cx=cx XOR wordmask= 0x0000 XOR 0x00CF=0x00CF.)</em>
<em />
<em>movl operand, %eax # eax=0x000000AA. ( copy the 32-bit value of operand=0x000000AA to eax.)</em>
<em>movl %eax, %ebx # ebx=0x000000AA.( copy the value of eax to ebx.)</em>
<em>movl %eax, %ecx # ecx=0x000000AA. ( copy the value of eax to ecx.)</em>
<em />
<em>shll $3,%eax # eax=0x00000550. ( logical shift left of eax=0x000000AA by 3 bits.)</em>
<em />
<em># 0x000000AA=00000000000000000000000010101010</em>
<em />
<em># =>00000000000000000000010101010000=0x00000550.</em>
<em />
<em>rorl $4,%ebx # ebx=0xA000000A. (Rotate right ebx=0x000000AA by 4 bits.)</em>
<em />
<em># 0x000000AA=00000000000000000000000010101010</em>
<em />
<em>3 =>10100000000000000000000000001010= 0xA000000A.</em>
<em />
<em>sarl %ecx # here the count value to shift the bits not mentioned.</em>
<em />
<em># It is the arithmetic shift right instruction. the shifted left bits filled with MSB bit.</em>
See more about logic program at brainly.com/question/14970713
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