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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Probably better for the comparisson: cmp edx, 0 jne endif

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it going to loop until the number is eqqual or less than 1.Or it will only show register output once?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That depends on the initial value of x and how you implement the while loop. The code I posted above will not loop.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if i am using the while loop everything will be like this(L1:................loop L1) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What do you mean? Do you have some ASM?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For me, it gets into an infinite loop because of the jnz L1. You should probably update %eax somewhere.
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