Problem 1
---------
goose:
        pushl   %ebp
        movl    %esp, %ebp
        movl    $-4, %eax
        popl    %ebp
        ret
        .size   goose, .-goose

cow:
        pushl   %ebp
        movl    %esp, %ebp
        movl    8(%ebp), %eax
        subl    12(%ebp), %eax
        popl    %ebp
        ret
        .size   cow, .-cow
.globl pig
        .type   pig, @function
pig:
        pushl   %ebp
        movl    %esp, %ebp
        movl    8(%ebp), %eax
        leal    (%eax,%eax,2), %eax
        popl    %ebp
        ret
        .size   pig, .-pig
.globl sheep
        .type   sheep, @function
sheep:
        pushl   %ebp
        movl    %esp, %ebp
        movl    8(%ebp), %eax
        shrl    $31, %eax
        popl    %ebp
        ret
        .size   sheep, .-sheep
.globl duck
        .type   duck, @function
duck:
        pushl   %ebp
        movl    %esp, %ebp
        pushl   %ebx
        movl    8(%ebp), %ebx
        pushl   %ebx
        call    sheep
        movl    %ebx, %edx
        testl   %eax, %eax
        je      .L6
        negl    %edx
.L6:
        movl    %edx, %eax
        addl    $4, %esp
        popl    %ebx
        popl    %ebp
        ret
        .size   duck, .-duck
.globl goose
        .type   goose, @function

Problem 2
---------
a b: which flags are set (1), rest are clear (0)
-4 0xfffffffc: ZF
4 0xfffffffc: SF
-1 1: CF
2 0x8000000: OF
0x7fffffff 0x80000000: OF
0x8000000 0x7fffffff: SF CF OF
1 0x7ffffffff: none
0x80000000 0x80000000: ZF
0x7fffffff 0xffffffff: SF


Multi-word arithematic, ie, implementing long longs

Problem 3
---------
fun2

Problem 4
---------
Fragment 3

Problem 5
---------
int dog(int x, int y) {
  int i, result;

  result = 1;
  for(i=x;i< y;i+=2)
    result = result*i;
  return result;
}

Problem 6
---------
int cat(int x, int y) {
  int i, ret;

  ret = 0;
  i = x;
  while(i--)
  {
    ret = ret+i*y;
  }
  return ret;
}

Problem 7
---------
int frog(int a, int b) 
{
  int result;

  switch(a)
  {
   case 1:
   case 5:
     result = 7;
     break;
   case 3:
     result = b-1;
     break;
   case 6:
     b++;
   case 7:
     result = b;
     break;
   default:
    result = -1;
  }
  return result;
}

Problem 8
---------
  ret address    0x04
  saved ebp      0x00
  len	      0x08
  buf            0xf8
  esp            0xf0
  pointer to ``\%s'' 0xf0

0x696e6f74


(gdb) x 0xbffff990
0xbffff990:     0x00000008
(gdb) x 0xbffff98c
0xbffff98c:     0x08048584
(gdb) x 0xbffff988
0xbffff988:     0xbffff900 {\em buffer overwrite from null character!}
(gdb) x 0xbffff984
0xbffff984:     0x736e656b
(gdb) x 0xbffff980
0xbffff980:     0x63696863
(gdb) x 0xbffff97c
0xbffff97c:     0xbffff980
(gdb) x 0xbffff978
0xbffff978:     0x08048640


Problem 9
---------
bear:
        pushl   %ebp
        movl    %esp, %ebp
        pushl   %esi
        pushl   %ebx
        movl    8(%ebp), %edx
        movl    12(%ebp), %ebx
        movl    16(%ebp), %esi
        cmpl    %ebx, %edx
        jge     .L7
        movl    global, %ecx
.L5:
        movl    %esi, %eax
        imull   %edx, %eax
        leal    (%ecx,%eax), %ecx
        incl    %edx
        cmpl    %ebx, %edx
        jl      .L5
        movl    %ecx, global
.L7:
        movl    global, %eax
        popl    %ebx
        popl    %esi
        popl    %ebp
        ret

Problem 10
----------
0xbffff9a4	0x3	n
0xbffff9a0	0x21	val
0xbffff99c	0x80483db	return address
0xbffff998	0xbffff9a8	saved ebp
0xbffff994	n/a		saved ebx
0xbffff990	0x2		n-1		
0xbffff98c	0x21		val
0xbffff988	0x80483b3	return address
0xbffff984	0xbffff998	saved ebp
0xbffff980	0x21		saved ebx (haven't done this yet..)
0xbffff97c	


the lea instruction (0x80483b3