Activation Records and Code Generation in Compilers - Prof. Sean P. Peisert, Study Guides, Projects, Research of Computer Science

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Code Generation

Lecture 21 Dr. Sean Peisert – ECS 142 – Spring 2009 1

Status

1 week to go on project 3

Read Sec. 7.1 through 7.4 by Monday, May 18 2

The Activation Record

The stack discipline guarantees that on function exit $sp is the same as it was on function entry

No need to save $sp

We need the return address

It’s handy to have a pointer to start of the current activation

This pointer lives in register $fp (frame pointer)

Reason for frame pointer will be clear shortly 4

The Activation Record

Summary: for this language, an AR with the caller’s frame pointer, the actual parameters, and the return address suffices.

Picture: consider a call to f(x,y), the AR will be: 5 old FP y x High Addresses

FP

SP

AR of f Low Addresses

Calling Sequence for f(x,y) 7

FP

SP

old FP y x

FP

SP

FP

SP

FP

SP

old FP y x return Before call On entry Before exit After call

Code Generation for Variables 8

Variable references are the last construct

The “variables” of a function are just its parameters

They are all in the AR

Pushed by the caller

Problem: Because the stack grows when intermediate results are saved, the variables are not at a fixed offset from $sp

Code Generation for Variables 10

Example: For a function def f(x 1 ,x 2 ) = e the activation and frame pointer are set up as follows: FP SP old FP x 2 x 1 return x 1 is at fp + 4 x 2 is at fp + 8 Thus: where z = 4 * i cgen(xi) = lw $a0 z($fp)

Summary

The activation record must be designed together with the code generator

Code generation can be done by recursive traversal of the AST

We recommend you use a stack machine for your Cool compiler (it’s simple) 11

Stack Machine Pros & Cons

The stack machine has activation records and intermediate results interleaved on the stack

Advantage: very simple code generation

Disadvantage: Very slow code

Storing/loading temporaries requires a store/load and $sp adjustment 13 AR Intermediates AR Intermediates

A Better Way

Keep Temporaries in the AR

Code generator must assign a location in the AR for each temporary. 14

How Many Temporaries?

• Let NT(e)= # of temps needed to evaluate e
• NT(e 1 + e 2 )

Needs at least as many temporaries as NT(e 1 )

Needs at least as many temporaries as NT(e 2 ) + 1

Space used for temporaries in e 1 can be reused for temporaries in e 2 16 

The Equations

NT(e 1 + e 2 ) = max(NT(e 1 ), 1 + NT(e 2 )) NT(e 1 - e 2 ) = max(NT(e 1 ), 1 + NT(e 2 )) NT(if e 1 = e 2 then e 3 else e 4 ) = max(NT(e 1 ), 1 + NT(e 2 ), NT(e 3 ), NT(e 4 )) NT(id(e 1 ,…,en) = max(NT(e 1 ),…,NT(en)) NT(int) = 0 NT(id) = 0 Exercise: Is this bottom-up or top-down? Exercise: What is NT(…code for fib…)? 17

Picture

19 Old FP xn ... x 1 Return Addr. Temp NT(e) ... Temp 1

Revised Code Generation 20

• Code generation must know how many temporaries to use at each point
• Add a new argument to code generation: the position of the next available temporary.