Question

After running the program below, what will be the values in registers EAX and EBX?

mov EAX,0123ABCDh

      mov EBX,4455AABBh

      PUSH BX

      PUSH AX

      POP AX

      POP BX

Question 2 options:

	EAX=ABCD 0123 EBX=AABB4455
	EAX= ABCD0123 EBX=4455AABB
	EAX=0123ABCD   EBX=4455AABB
	EAX=ABCD0123    EBX=AABB4455

Answer 1

Correct Answer:

Explanation:

A stack is a linear data structure in which insertion and deletion operation performed at the top always.

The first two statements load the register EAX and EBX as given below:

EAX = 0123ABCD

EBX = 4455AABB

The register AX and BX have the lower 16 bits of the register EAX and EBX respectively as given below:

AX = ABCD

BX = AABB

The PUSH operation will push the content of the register BX onto the stack, the second PUSH operation push the content of the register AX onto the stack, and the status for the stack after the third instruction will be as given below:

PUSH AX

The fourth instruction POP operation will pop the top element from the stack and load the register AX.

POP AX

The fifth instruction POP operation will pop the top element from the stack and load the register BX.

POP BX

Now, the stack will be empty and the content in the register AX and BX will be as given below:

AX = ABCD

BX = AABB

So, the status of the register EAX and EBX will be as given below:

EAX = 0123ABCD

EBX = 4455AABB

So, only the third option is correct.