Assembly Language 8086 INFORMATION: Assume when we start ESP has the value $2000

Question

Assembly Language 8086

INFORMATION:

Assume when we start ESP has the value $2000

MOVL $0X34EF, %EAX

MOVL $0X55, %EBX

MOVL %EBX, %ECX

ADDL $8, %ECX

MOVL %ECX, %EDX

PUSHL %EAX

PUSH           %EBX

PUSH           %ECX

PUSH           %EDX

MOVL $0, %EAX

MOVL $5, %EBX

MOVL $9, %ECX

MOVL $7, %EDX

This is what happens I think:

ESP is the stack pointer.

A = 34EF

B = 55

C = EB

mov B to C, so c = 55

Add 8 with C so 55 + 8 = 63

mov c,d so d = 63

push a so stack will have 34EF esp = 2000

push b , so stack will have 34EF, 55 esp = 2001

push c stack = 34EF, 55, 55, esp = 2002

push d 34EF, 55, 55, 63 esp = 2003

mov 0 to a, so a=0

mov 5 to b, so b = 5

mov 9 to c, so c = 9

mov 7 to d, so d = 7

QUESTION:

1) If we want to retrieve the original values back into the registers, list the instructions needed.

2) Write a procedure to add the value of EAX 10 times to itself. Use a loop inside the function. Use register ECX as a loop counter. Write the main program to place 5 into EAX and 2 in ECX and then call the function. After the return of the function the value of ECX must be 2 again but the value of EAX must be 50.

Explanation / Answer

1. for the first part

POP %EDX

POP %ECX

POP %EBX

POP %EAX

2. MOV %EAX ,$05;

MOV %ECX ,$02;

PUSH %ECX;
Label_X: ADD %EAX,$05;
ADD %ECX,$02;

CMP %ECX, 20;

JL Label_X;

POP %ECX;

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