CODEWITHSUNDEEP
assembly
user3483844
user3483844

Reputation: 143

Assembly code of a loop?

I'm having some issues understanding some of the parts of the assembly code I am suppose to decipher into a loop

prob2:
pushl  %ebp
movl   %esp, %ebp
movl   8(%ebp), %eax
cmpl   $1, %eax
je     .L1
.L6:
testb  $1, %al
je     .L3
leal   1(%eax,%eax,2), %eax
jmp    .L4
.L3:
shrl   %eax
.L4:
cmpl   $1, %eax
jne    .L6
.L1:
popl   %ebp
ret

Now I am given the general outline of the C code

void prob2(unsigned n)
{
while (________________) {
if (_______________) {
      ______________;
} else {
 _______________;
}
}
}

I've figured out that the while loop tests to see whether n < 1, but in the next if-else part I am confused as to what the %al part is. I assume it's testing to see if it is equal to 1 and if it is (which it never will be because the while loop breaks when n = 1) then it shifts n to the right by a byte else it does the leal part and compares to see if it's not equal to 1 yet.

Does that sound right?

Upvotes: 2

Views: 130

Answers (1)

Scott Hunter
Scott Hunter

Reputation: 49803

%al is lowest byte of the register %eax, which is where you have determined that n is being kept (based on your interpretation of cmpl $1, %eax).

Upvotes: 1

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