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Discussion Section – 11/3/2012

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Discussion Section – 11/3/2012

Midterm review

- A bottom up parsing example
- A Type Unification Example
- Run Time Memory Subdivision
- A General Activation Record
- An illustration of implementing recursion using stacks
- Examples

- Grammar:
S-> CC

C->cC

C->d

Sand C are non-terminals

cand d are terminals

Show the stack trace

for the grammar:

cdccd

Parsing Table (courtesy: Dragon Book)

1. Simple:

deff(x):

return -x

y = f(1)

2. Parametric Polymorphism

deff(z): return z

x = f(0)

y = f("hello")

3. Restrictions?

deff():

defg(x): return x

q = g([])

r = g(3)

- An illustration of implementing recursion using stacks

Objective Caml language:

let binom n k = ...

let test x y =

let a = binom x y in

let b = binom x (y+1) in

a + b (* Return the sum *)

- If we compile this code and then disassemble the result we get the following (using Intel syntax):
- On calling test,
sub sp,12

mov *sp+4, eax

mov *sp, ebx

Call binom

Mov *sp+8, eax

Movebx,*sp

Add ebx, 1

Moveax,*sp+4

Call binom

Mov *sp+8,ebx

Lea eax, [eax+ebx-1]

Add *sp,12

return

- Describe the calling conventions for binom: Where are parameters n and k stored? Where is the result located on return?
- Draw a diagram showing the layout of the stack and the register values right before the second call to binom. Your diagram should show where each argument and local variable is stored.

- Describe the calling conventions for binom: Where are parameters n and k stored? Where is the result located on return?
- The argument n is passed in the eax register, and k in ebx. We can tell this because eaxhas the same value both times binom is called, but the second time ebx's value is incremented. The return value is passed in eax, which we can tell because we can trace the two values added for a + b back to the values of eax right after the two calls to binom.

- Draw a diagram showing the layout of the stack and the register le right before the second call to binom. Your diagram should show where each argument and local variable is stored.
- Stack (growing downward) and Registers

- 080483b4 <test>:
- 80483b4: push ebp ; esp -= 4; *esp = ebp
- 80483b5: movebp,esp
- 80483b7: sub esp,0x18
- 80483ba: movDWORD PTR [ebp-0x14],ecx
- 80483bd: movDWORD PTR [ebp-0x18],edx
- 80483c0: moveax,DWORD PTR [ebp-0x14]
- 80483c3: movedx,DWORD PTR [eax]
- 80483c5: moveax,DWORD PTR [ebp-0x18]
- 80483c8: moveax,DWORD PTR [eax]
- 80483ca: imuledx,eax ; edx = edx * eax
- 80483cd: moveax,DWORD PTR [ebp-0x14]
- 80483d0: movecx,DWORD PTR [eax+0x4]
- 80483d3: moveax,DWORD PTR [ebp-0x18]
- 80483d6: moveax,DWORD PTR [eax+0x4]
- 80483d9: imuleax,ecx
- 80483dc: lea eax,[edx+eax*1] ; eax = edx + eax
- 80483df: movDWORD PTR [ebp-0x4],eax
- 80483e2: moveax,DWORD PTR [ebp-0x4]
- 80483e5: leave ; esp = ebp; pop ebp
- 80483e6: ret
- This function takes two parameters, passed in registers ecx and edx respectively. Its result is returned
- in register eax. Decompile (translate) this assembly into equivalent C code. Hint: This code implements a
- well-known mathematical operation.

- Solution : Dot Product