Question

The following code fragment is expressed in arm assembly code. Fill in the blanks, so that it is equivalent to the following C code.

int counter;

int x = 5;

int y = 6;

for (counter =10; counter >0;counter--)

IF(X==Y)

Y = Y + 1 ;

ELSE

Y = Y + 2}

Fill in the blanks in the following code:

MOV__________ ;loop counter into r0-ten times round the loop

MOV__________ ;Value of y loaded into r1

MOV__________ ;Value of x loaded into r2

Next CMP ____________ ;assume r1 contains y and r2 contains x: compare them

BNE ____________ ;if not equel then branch to the else part

ADD ____________ ;if equal fall through to here and add one to y

B _____________ ;now skip past the else part

plus2 ______________ ;ELSE part add 2 to y

counter ______________ ;decrement loop counter

BNE _____________ ;continue until all done

Answer 1

C language code:

#include<stdio.h>

int main() {
int counter;

int x = 5;

int y = 6;

for (counter =10; counter >0;counter--)
{
if(x==y)
{
y=y+1;
}
else
{
y=y+2;
}
}
}

-----------------------------------------------------------

Ans

Here is the equivalent assembly code:

Ans here is the equivalent assembly code:

. file "Test.c"

.def ___main; .scl 2; .type 32; .endef

.text

.globl _main

.def _main;. scl 2; type 32; .endef

_main:

pushl %ebp

movl %esp, %ebp

andl $-16, %esp

subl $16, %esp

call ___main

movl $5, 4(%esp)

movl $6, 8(%esp)

movl $10, 12(%esp)

jmp L2

L5:

movl 4(%esp), %eax

cmpl 8(%esp), %eax

jne L3

addl $1, 8(%esp)

jmp L4

L3:

addl $2, 8(%esp)

L4:

subl $1, 12(%esp)

L2:

cmpl $0, 12(%esp)

jg L5

movl $0, %eax

leave

ret

.ident "GCC: (tdm-1) 4.9.2"

_________________________________________________________

I didn't filled the blanks because different system generates different address spaces.