IA32 Assembly

My question is geared towards the IA32 Assembly language that a C program produces. I've been given the following and my task is to figure out what the funciton does. I've written my comments next to each line and as far as I can tell it looks to me that if local variable a >= y then 0 gets returned, if not then other actions are taken until a is > and then a value for b is returned. Any help on exactly what this function does would be greatly appreciated.

Code:

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pushl  %ebp
        movl    %esp, %ebp
        subl    $16, %esp
        movl    $0, -8(%ebp)  SET LOCAL VARIABLE B TO 0
        movl    $0, -4(%ebp)  SET LOCAL VARIABLE A TO 0
        jmp    .L2              JUMP TO L2

                        IF (A<Y)
.L3:
        movl    -4(%ebp), %eax  MOVE A TO EAX
        sall    $2, %eax        A * 2^2 IN EAX
        addl    8(%ebp), %eax  ADD X + A IN EAX
        movl    (%eax), %eax        MOVE (X + A) IN EAX
        addl    %eax, -8(%ebp)  ADD (X + A) IN B
        addl    $1, -4(%ebp)        ADD 1 TO A
.L2:
        movl    -4(%ebp), %eax  MOVE A TO EAX
        cmpl    12(%ebp), %eax        COMPARE A TO Y
        jl      .L3                IF A < Y JUMP TO L3
        movl    -8(%ebp), %eax  MOVE B TO EAX
        leave
        ret                        RETURN B

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I won't give you the answer directly, because this sounds like homework. However, I'll give you a starting point, because you may not necessarily be looking at this the right way. Or maybe you are, and this starting point is not very helpful; at any rate, see what you can do if I tell you the prototype for this function is:

Code:

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int f(int *array, int size);

---

That's clearly not very well optimized...

Code:

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        push  %esi
        push  %edi
        mov    $12(%esp), %esi
        mov    $16(%esp), %edi
 
        movl    $0, %ecx  SET LOCAL VARIABLE B TO 0
        movl    $0, %edx  SET LOCAL VARIABLE A TO 0
        jmp    .L2              JUMP TO L2

                        IF (A<Y)
.L3:
        addl    %ecx, (%esi), %eax       
        addl    %esi, 4
        incl      %edx
.L2:
        cmpl    %edi, %edx        COMPARE A TO Y
        jl      .L3                IF A < Y JUMP TO L3
        movl    %ecx, %eax  MOVE B TO EAX
        pop    %edi
        pop    %esi
        ret                        RETURN B

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I actually find this easier to understand...

--
Mats

Quote:

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Originally Posted by cas

I won't give you the answer directly, because this sounds like homework. However, I'll give you a starting point, because you may not necessarily be looking at this the right way. Or maybe you are, and this starting point is not very helpful; at any rate, see what you can do if I tell you the prototype for this function is:

Code:

---

int f(int *array, int size);

---

---

It is homework and I'm not looking for the exact answer I just need a big push in the right direction....so this a function that fills an array? Something like with for loop below?

int f(int *array, int size)

for (int a = 0; a < size; a++)

Yes.

--
Mats

last question....I read the sall $2, %eax as 4 * a where a is initially = 0. So my I would say the the array just fills with multiple's of 4 up to whatever size is specified. Does that sound right?

something like (0, 4, 8, 12,......)

It is not storing data into the array. But yes, it walks in steps of 4.

--
Mats

Quote:

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Originally Posted by matsp

It is not storing data into the array. But yes, it walks in steps of 4.

--
Mats

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eh...I'm still confused...if it's not storing data into an array what is it doing?

Quote:

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Originally Posted by Smohr

eh...I'm still confused...if it's not storing data into an array what is it doing?

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Code:

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movl    (%eax), %eax

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Figure out what that means and it might start falling into place.

If you're at your wits' end, write something simple (that is similar to what you expect this code to be doing) in C and see what gcc spits out for the assembly. Perhaps seeing how gcc translates C->asm will help figure out what it did here.

Quote:

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Originally Posted by cas

Code:

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movl    (%eax), %eax

---

Figure out what that means and it might start falling into place.

If you're at your wits' end, write something simple (that is similar to what you expect this code to be doing) in C and see what gcc spits out for the assembly. Perhaps seeing how gcc translates C->asm will help figure out what it did here.

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I read that as move the value that's store in the memory location located in eax into eax

so eax = *eax

sound right?

Quote:

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Originally Posted by Smohr

I read that as move the value that's store in the memory location located in eax into eax

so eax = *eax

sound right?

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Yep. Now you should understand why it's not storing data in the array...

Quote:

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Originally Posted by cas

Yep. Now you should understand why it's not storing data in the array...

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Nope...still lost

Quote:

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Originally Posted by Smohr

Nope...still lost

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Code:

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int main(void)
{
  int array[1];
  int i = 0;

  /* Which of these is storing a value in the array? */
  *array = i;
  i = *array;
}

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Which compares to what the assembly version is doing?

Quote:

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Originally Posted by cas

Code:

---

int main(void)
{
  int array[1];
  int i = 0;

  /* Which of these is storing a value in the array? */
  *array = i;
  i = *array;
}

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Which compares to what the assembly version is doing?

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*array = i