Question

Assembly Code

may I get an explanation for the 3 parts in this question

Given that . Ar is a label at address 510 EAX has the value 139 . EBX has the value 102 Which bytes in memory will be accessed by the instruction movl $Ar + 38, %ecx Write your answer in the form startByteEndByte with no space in between. If memory is NOT accessed write -1 for your answer. For example, if your answer was bytes 527 832 write your answer as 527832. Or if you answer was bytes 56-328, write your answer as 56328. If only a single byte is read from memory, still write your answer as startByteEndByte. For example, if only byte 112 was read from memory write your answers as 112112 Given that . Ar is a label at address 103 . EAX has the value 558 . EBX has the value 362 Which bytes in memory will be accessed by the instruction Write your answer in the form startByteEndByte with no space in between. If memory is NOT accessed write -1 for your answer. For example, if your answer was bytes 527 832 write your answer as 527832. Or if you answer was bytes 56-328, write your answer as 56328. If only a single byte is read from memory, still write your answer as startByteEndByte. For example, if only byte 112 was read from memory write your answers as 112112 Given that Ar is a label at address 521 EAX has the value 154 EBX has the value 456 Which bytes in memory will be accessed by the instruction movb Ar + 531%eax, %ebx), %cl Write your answer in the form startByteEndByte with no space in between. If memory is NOT accessed write -1 for your answer. For example, if your answer was bytes 527 832 write your answer as 527832. Or if you answer was bytes 56-328, write your answer as 56328. If only a single byte is read from memory, still write your answer as startByteEndByte. For example, if only byte 112 was read from memory write your answers as 112112.

Answer 1

1. movl $Ar +38, %ecx

in above instruction movl trying to mov a long value into  general purpose register ecx

here $Ar represent the value of Ar and  $Ar +38 will return a value while adding 38 (a 32 bit number) into value of Ar

As the base address Ar=510 and 38 is a value which is goinig to be added in value of Ar , therefore the address will be same as 510 and only value will get updated .

Ans: 510510 i.e. no change in address

2. movw Ar,%cx

since movw just copy the value from Ar to register cx and therefore no memory access is needed.

Ans: -1 i.e. no excess of address is needed

3. movb Ar + 53(%eax,%ebx), %cl

here (%eax,%ebx) means (address of eax + address of ebx) that is 154+456=600

Now + 53(%eax,%ebx) or theoretically   signed-offset value(address of eax + address of ebx)

means 600+ offset value = 600+53=653

now Ar will provide base address and (here StartByte) and address value result of   +53(%eax,%ebx) will give end address (here EndByte)

Ans. 521653