Activation Records and Frames in Compiler Design, Slides of Compilers

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COMPILERS

Activation Records

Subprogram Invocation Mechanics

 Save status of caller.

 Process parameters.

 Save return address.

 Jump to called subprogram.

 … do stuff ...

 Process value-result/result parameters and

function return value(s).

 Restore status of caller.

 Jump back to caller’s saved position.

Stack-based Recursion

 When recursion is implemented using a

stack, activation records are pushed onto

the stack at invocation and popped upon

return.

 Example:

int sum ( int x )

if (x==0) return 0;

else return (x + sum (x-1));

void main ()

{ sum (2); }

Recursion Activation Records

mainARI

retvalue (?) parm (x=0) dynamiclink staticlink return (sum) retvalue (?) parm (x=1) dynamiclink staticlink return (sum) retvalue (?) parm (x=2) dynamiclink staticlink return (main) mainARI

retvalue (?) parm (x=1) dynamiclink staticlink return (sum) retvalue (?) parm (x=2) dynamiclink staticlink return (main) mainARI

retvalue (?) parm (x=2) dynamiclink staticlink return (main)

sum(2)

main

sum(2)

main

sum(1)

sum(2)

main

sum(1)

sum(0)

Non-local Reference Example

 Example:

main { int x; sub SUBA { sub SUBB { x = 1; } SUBB; } sub SUBC { int x; int y; SUBA; } SUBC; }

breakpoint

breakpoint

breakpoint

breakpoint

Static Chains

local (x)

dynamiclink staticlink return (A) dynamiclink staticlink return (C) local (x) local (y) dynamiclink staticlink return (main) local (x)

dynamiclink staticlink return (C) local (x) local (y) dynamiclink staticlink return (main) local (x)

local (x) local (y) dynamiclink staticlink return (main)

SUBC

main

SUBC

main

SUBA

SUBC

main

SUBA

SUBB

breakpoint1 breakpoint2 breakpoint

Static Chains vs. Displays

 Static chains require more indirect

addressing – displays require a fixed

amount of work.

 Displays require pointer maintenance on

return – static chains do not.

 Displays require “backing up” of display

pointer – static chains require static links

in each activation record.

Dynamic Scoping

 Dynamically scoped languages can be

implemented using:

 Deep Access

 Follow the dynamic chains to find most recent

non-local name definition.

 Shallow Access

 Maintain a separate stack for each name.

Frame Pointers

 Stack frames are usually supported by:

 stack pointer - points to top of stack

 frame pointer - points to top of previous frame

frame

pointer

stack

pointer

AR g

AR h

frame

pointer

stack

pointer

AR g

AR f AR f

After call to h()

View Shifts

 On a Pentium machine,

 M[SP + 0] <-- FP

 (^) save old frame pointer

 FP <-- SP

 (^) move frame pointer to top of stack

 SP <-- SP - K

 (^) move stack pointer to end of new frame

 On machines which use registers for frame

optimisation, remember to save registers

in temporary variables.

Parameter Passing

 Registers are more efficient than copying

every parameter to the stack frame.

 Registers are limited so pass first k parameters

in registers and rest in frame.

 Nested subprogram calls require saving

and restoring so there is dubious cost

savings!

 leaf procedures, different registers, done with

variables, register windows

 How does C support varargs?

Return Addresses

 Traditionally a stack frame entry.

 More efficient to simply use a register.

 Same saving procedure necessary as before for

non-leaf subprograms.

Frame Implementation 1/

 A Frame class corresponds to the frame

for each subprogram.

 During translation, frames are created to track

variables and generate prologue/epilogue

code.

 Frame can be an abstract class with

instantiations for different machine

architectures.

 Each instantiation must know how to

implement a “view shift” from one frame to

another.

Frame Implementation 2/

 Each time a local variable is defined, a

method of Frame can be called to allocate

space appropriately (on stack frame or in

registers).

 f.allocLocal (false)

 Parameter indicates if variable requires

memory (escapes) or not - should we allocate

stack space or temporary?

 Allocating a temporary for each variable

can be slow - future stages will optimise by

reusing both registers and space.