Question

what will be the final values of cx and dx when the following code executes? .data array sword 8,2,3,5,-4,6,0,4 .code mov cx,1 mov esi,2 mov ax,array[esi] mov bx,array[esi 4] cmp ax,3 jae l2 cmp bx,4 jb l1 jmp l3 l1: mov cx,4 l2: mov dx,5 jmp l4 l3: mov dx,6 l4:

Answer 1

In the given assembly code, the value of CX remains 1 and the value of DX is set to 5 after execution.

Step-by-step explanation:

The provided code snippet appears to be written in x86 assembly language and performs a series of operations on an array of signed words (16-bit integers).

The code follows these steps:

• mov cx,1 - Initializes register CX to 1.
• mov esi,2 - Sets the Source Index (ESI) to point to the third element of the array.
• mov ax,array[esi] - Moves the value at ESI (which is the third element, 3) into AX.
• mov bx,array[esi*4] - Moves the value at ESI times 4 (which would be the value at the 9th position but the array only has 8 elements, resulting in an error).
• cmp ax,3 - Compares value in AX (which is 3) with 3.
• jae l2 - Since AX is equal to 3, the jump to label l2 is taken.
• At l2, mov dx,5 - Register DX is set to 5.

Since there is a jump to label l4 following l2, the code bypasses l3 completely and the final value of register CX remains unchanged from its initial value of 1, while DX ends up with the value of 5.