Can someone explain me the output values for the below assembly code

Question

Can someone explain to me the output values for the below assembly code. Like how can I arrive at the solution?

Arr Byte 5D, 2E, 97, 4B, A5 , 7E , 84 , 9A , 61 , 8B 

     mov ecx, 5
     mov esi, offset Arr 

LI:  rcr array[esi], 1 
     Addi esi,esi, 2 
     Loop LI

`Addi esi, esi, 2` does not look like an x86 command. I assume `add esi, 2` or maybe `lea esi, [esi+2]`. Otherwise it looks like rotating some bytes in array though carry flag. — Alex Guteniev, Dec 28 '21 at 19:09
@fuz, I've read about y86, it is simplified ISA that doesn't have complex x86 stuff like `rcr` or `loop` — Alex Guteniev, Dec 29 '21 at 12:28

Alex Guteniev · Accepted Answer · 2021-12-29T12:27:11.833

Let's replace Addi esi,esi, 2 with lea esi, [esi+2] because:

There's no Addi in x86
The intention seem to be to increment esi by 2
add command affects CF, and rcr presumably expects CF from the previous iteration

No idea if there's some assembler to accept Addi.

The whole loop then runs in even-indexed elements of Arr (assume indexing is zero-based)

RCR rotates bits right through CF (Carry Flag). that is out of eight bits of each element each shifted towards less significance, the least significant bit is put into CF and the most significant bit is extracted from CF. See also:

Knowing that, you can compute the new values of Arr element, except the 0th element: you don't know the initial value of CF.

Can someone explain me the output values for the below assembly code

1 Answers1