Question

Assembly baisics
Can somebody explain how to solve these questions?

Answer 1

I am a noob at assembly, but for the second one is that not equivalent to
R0=R0+0x200
which would *not* be the same as R0=R0+0x2000000

Answer 2

I think bitfill will be at more significant places, so instead of 0x20000000, it becomes 0x00000200.

Answer 3

Good point rbx, id like to verify that.but for number 8, where do i find the content to add 0x200 too?

Answer 4

Yes, turingtestt, i believe you are correct but not sure what value to use for the original value for r0

Answer 5

it says the content is 0x4F, so I figure we are putting 0x4F+0x200 (no bitfill) into register R0

Answer 6

oh no, I guess it's the other value, so 0x2F280000 + 0x00000200 into R0

Answer 7

my understanding is to read it from right to left

Answer 8

Sorry, I don't understand much but ins't 0x4F the value in memory ?

Answer 9

the heading is strange, in that it has "register=value in hex" and "content" as separate columns

Answer 10

Thats what i dont get, where i am looking for the content, in the register or in memory. If in memory, howdo i find that from the info in the register? I am confussed as to what vakuesi am using

Answer 11

I think I am getting some answers after reading this http://www.rz.uni-karlsruhe.de/rz/docs/VTune/reference/INST_STRH.htm

Answer one thing if you can. Does STRH R0,[R1] means that content in R0 is stored at the memory location specified by the contents of R1?

Answer 12

If i take time to reply please forgive me, im at work till 6

Answer 13

If yes, then 8) 0x2F280200
and all the answers to question 9 are correct.

Answer 14

I believe that is how the command works, brackets indicate a location

Answer 15

Do you need any explanation regarding this, because the answers are correct ?

Answer 16

It would be great if you could... will you be on here tonight around 7?

Answer 17

Sorry, that would hardly be possible, it is 3 A.M. here right now. I can try explaining what I understand before I log off.

Answer 18

Thanks man, much abliged

Answer 19

8) 0x2F280000 + 0x00000200 = 0x2F280200
9)a) R0 remains unchanged, R0 = 0x2F280000
9)b) R1 remains unchanged, R1 = 0x20002000
9)c),d),e) No changes to those memory locations.
9)d) STRH R0,[R1] stores contents of R0 (0x2F280000) to the address given in R1 (0x20002000). Therefore the value at memory location 0x20002000 becomes what was stored in R0 = 0x2F280000.

Answer 20

Thanks, i apreciate all your time, i am reviewing your work now my savior.