After running this Assembly language program below, give the final values in EAX, EBX, and ECX. MOV EAX, MOV EBX, MOV ECX, PUSH AX PUSH BX PUSH CX PUSH BX PUSH AX PUSH CX POP EAX POP EBX POP ECX (Points : 10)

1.EAX= EBX= ECX=

2.EAX= EBX= ECX=

3.EAX= EBX= ECX=

4.EAX= EBX= ECX=

5.EAX= EBX= ECX=

See explaination please.

Explanation:

We are going to get a solution by making use of the instruction;

Given The instruction,

MOV EAX,

MOV EBX,

MOV ECX,

This is going to move the defined values, say , , , to the registers EAX, EBX, ECX, respectively

Now,

PUSH AX

This push instruction pushes the first 8 bits in register EAX, to a stack,

again, PUSH BX

Also this push instructions will also push the first 8 bits in register EBX, to the stack

PUSH CX

This push instruction pushes the first 8 bits in register ECX, to the stack,

similarly PUSH BX PUSH AX PUSH CX pushes the 8 bit registers in the same stck,

NOw, the instructions POP EAX POP EBX POP ECX

will lead to the contents of the respective registers as

EAX= EBX= ECX=

See explaination please.

Explanation:

We are going to get a solution by making use of the instruction;

Given The instruction,

MOV EAX,

MOV EBX,

MOV ECX,

This is going to move the defined values, say , , , to the registers EAX, EBX, ECX, respectively

Now,

PUSH AX

This push instruction pushes the first 8 bits in register EAX, to a stack,

again, PUSH BX

Also this push instructions will also push the first 8 bits in register EBX, to the stack

PUSH CX

This push instruction pushes the first 8 bits in register ECX, to the stack,

similarly PUSH BX PUSH AX PUSH CX pushes the 8 bit registers in the same stck,

NOw, the instructions POP EAX POP EBX POP ECX

will lead to the contents of the respective registers as

EAX= EBX= ECX=

Explanation:

b and c ( ip address and port number)