Question

how to write this in assembly
while(x>1)
{
if(x is even)
x=x/2
}

Answer 1

The basic structure of a while loop is also shown here: http://openstudy.com/study#/updates/5124d00be4b086b98ebdbc10

The loop body can be done in two ways: slow and faster. I'll explain the slow method in more details and I'll give some hints for the faster method.

For the slow method, you'll need a division. Divisions can be a bit complex to use. You'll need to put the dividend into two registers: the upper half in edx and the lower half in eax. Then you can use the div instruction with the divisor as argument. The quotient will be in eax and the remainder in edx.

So in code. Let's assume x is an unsigned int (so the upper half is all 0s) and is located in esi:

mov edx, 0
mov eax, esi
mov edi, 2 # div only works on registers and
# memory locations
div edi
cmp edx, 1
jnz endif # if the compare is zero, the number
# is even so we need to jump when
# it's not
mov esi, eax # eax is still the result of the
# previous division. so no need
# to redo it
endif: ...


For the fast method, you can check certain bits to see if a number is even/odd. Also, division by a power of 2 can be replaced with a fast instruction.

Answer 2

Probably better for the comparisson:
cmp edx, 0
jne endif

Answer 3

is it going to loop until the number is eqqual or less than 1.Or it will only show register output once?

Answer 4

That depends on the initial value of x and how you implement the while loop. The code I posted above will not loop.

Answer 5

so if i am using the while loop everything will be like this(L1:................loop L1) ?

Answer 6

What do you mean? Do you have some ASM?

Answer 7

YES

Answer 8

Yes, you can use the loop instruction, but make sure that the loop counter in in %ecx (you seem to check %eax at the start of the loop)

Answer 9

but it does not loop

Answer 10

For me, it gets into an infinite loop because of the jnz L1. You should probably update %eax somewhere.