Code Generation

Outline

• Stack machines

• Abstract assembly code

• A stack machine implementation example

Stack Machines

• A simple evaluation model

• No variables or registers

• A stack of values for intermediate results

• Each instruction:

  • Takes its operands from the top of the stack

  • Removes those operands from the stack

  • Computes the required operation on them

  • Pushes the result to the stack

Example of a Stack Machine Program

• Consider two instructions:

  • push i – place the integer i on top of the stack

  • add – pop the topmost two elements, add them and put the result back on the stack

• Example program to compute \(7 + 5\)
  push 7
push 5
add
  

Why Use a Stack Machine?

• Each operation takes operands from the same place and puts results in the same place

• This means a uniform compilation scheme

• And therefore a simpler compiler

Why Use a Stack Machine?

• Location of the operands is implicit; always on top of the stack

• No need to specify operands explicitly

• No need to specify the location of the result

• Instruction encoding is more compact than instructions with registers

• Many bytecode interpreters use a stack machine model, for example, Java and Python

Optimizing the Stack Machine

• The add instruction does three memory operations:

  • Two read operations and one write operation

  • The top of the stack is frequently accessed

• Idea: keep the top of the stack in a dedicated register (called the “accumulator”)

• The add instruction is now
  acc := acc + top
  which is only one memory operation

Stack Machine with Accumulator: Invariants

• The result of computing an expression is always placed in the accumulator

• For an operation \(op(e_1, \ldots, e_n)\), compute each \(e_i\) and then push the accumulator (the result of evaluating \(e_i\)) on the stack

• After the operation, pop \(n - 1\) values

• After computing an expression, the stack is as before

Stack Machine with Accumulator: Example

• Compute \(3 + (7 + 5)\) using an accumulator:

  Code	Accumulator	Stack
  acc := 3	3	\(\langle init \rangle\)
  push acc	3	3, \(\langle init \rangle\)
  acc := 7	7	3, \(\langle init \rangle\)
  push acc	7	7, 3, \(\langle init \rangle\)
  acc := 5	5	7, 3, \(\langle init \rangle\)
  acc := acc + top	12	7, 3, \(\langle init \rangle\)
  pop	12	3, \(\langle init \rangle\)
  acc := acc + top	15	3, \(\langle init \rangle\)
  pop	15	\(\langle init \rangle\)

From Stack Machines to Three-address Code

• The compiler generates code for a stack machine with an accumulator

• Here we use an abstract RISC assembly language for simplicity

• The generated assembly code simulates the stack machine instructions with instructions and registers

Notes

• A register is a fast-access untyped global variable shared by the entire assembly program

• An instruction is a primitive statement in assembly language that operates on registers

• A load-store architecture: bring values into registers from memory to operate on them.

Simulating a Stack Machine with Assembly

• The accumulator is kept in a register, we will call it acc

• The stack is kept in memory

• The stack grows towards lower addresses

• The address of the next location on the stack is kept in a register, we will call it sp for stack pointer

• Memory is accessed with load and store instructions

• Assume a machine word is 32-bits

• Assume an arbitrary number of registers named t1, \(\ldots\), tn

Sample Instructions

• Load word: load a 32-bit word from address \(register_1 + offset\) into \(register_2\)
  lw r1 offset(r2)

• Store word: store a 32-bit word in \(register_1\) at address \(register_2 + offset\)
  sw r1 offset(r2)

• Load immediate value
  li reg imm

• Add \(register_2\) and \(register_3\) and store the result in \(register_1\)
  add r1 r2 r3

Example

• The stack machine code for 7 + 5:

  acc := 7	li acc 7
  push acc	sw acc 0(sp)
  	li t1 -4
  	add sp sp t1
  acc := 5	li acc 5
  acc := acc + top	lw t1 4(sp)
  	add acc acc t1
  pop	li t1 4
  	add sp sp t1

A Small Language

• We will generalize the previous example to a simple language; a language with only integers and integer operations

• Grammar \[\begin{aligned} Program & \rightarrow Function \; Program\\ & \quad \vert \; Function\\ Function & \rightarrow id(Args) \; begin \; E \; end\\ Args & \rightarrow id, Args\\ & \quad \vert \; id\\ E & \rightarrow int\\ & \quad \vert \; id\\ & \quad \vert \; if \; E_1 = E_2 \; then \; E_3 \; else \; E_4\\ & \quad \vert \; if \; E_1 + E_2\\ & \quad \vert \; if \; E_1 - E_2\\ & \quad \vert \; id(E_1, \ldots, E_n)\\ \end{aligned}\]

A Small Language

• The first function definition \(f\) is the “main” function

• Running the program on input \(i\) means computing \(f(i)\)

• Example program: Fibonacci numbers:

  fib(x)
  begin
    if x = 1 then 0 else
    if x = 2 then 1 else fib(x-1) + fib(x-2)
  end

Code Generation Strategy

• For each expression \(e\) we generate assembly code that:

  • Computes the value of \(e\) in acc

  • Preserves sp and the contents of the stack

• We define a recursive code generation function \(cgen(e)\) whose result is the code generated for \(e\)

Code Generation for Constants

• The code to evaluate an integer constant simply copies it into the accumulator:
  \(cgen(int) =\) li acc \(int\)

• Note that this also preserves the stack, as required

Code Generation for Addition

Code Generation Notes

• The code for \(e_1 + e_2\) is a template with “holes” for the code that evaluates \(e_1\) and \(e_2\)

• Stack machine code generation is recursive

• The code for \(e_1 + e_2\) consists of code for \(e_1\) and \(e_2\) glued together

• Code generation can be written as a recursive descent of the AST (at least for arithmetic expressions)

Code Generation for Subtraction

• New instruction: subtract \(register_2\) and \(register_3\) and store the result in \(register_1\)
  sub r1 r2 r3

  \(cgen(e_1 - e_2) =\)
  \(\qquad cgen(e_1)\)	; acc := the value \(e_1\)
  \(\qquad\)sw acc 0(sp)	; push that value on the stack
  \(\qquad\)li t1 -4
  \(\qquad\)add sp sp t1
  \(\qquad cgen(e_2)\)	; acc := the value of \(e_2\)
  \(\qquad\)lw t1 4(sp)	; retreive the value of \(e_1\)
  \(\qquad\)sub acc t1 acc	; perform the subtraction
  \(\qquad\)li t1 4	; pop the stack
  \(\qquad\)add sp sp t1

Code Generation for Conditionals

• We need flow control instructions and labels

• A label is a symbolic name that indicates a point in the code that can be jumped to

• The code for \(e_1 + e_2\) consists of code for \(e_1\) and \(e_2\) glued together

• New instructions:

  • Branch to label if \(register_1 = register_2\)
    beq r1 r2 label

  • Unconditional jump to label
    jump label

Code Generation for If Then Else

Code Generation for Functions

• Code for function calls and function definitions depends on the layout of the activation record

• A simple activation record is sufficient for the example language

  • The result is always in the accumulator; there is no need to store the result in the activation record

  • The activation record holds the actual parameters; for \(f(x_1, \ldots, x_n)\) push the arguments \(x_1, \ldots, x_n\) onto the stack

• The stack machine invariants guarantee that on function exit the stack is the same as it was before the arguments got pushed

• We need the return address

• It is also convenient to have a pointer to the current activation; this pointer will be stored in the register fp (frame pointer)

Layout of the Activation Record

• For the example language, an activation record with the caller’s frame pointer, the actual parameters, and the return address is sufficient

• Consider a call to \(f(x,y)\), the activation record would be:

Code Generation for a Function Call

• The calling sequence is the instructions (of both caller and callee) to set up a function invocation

• New instructions:

  • Jump to label and save the address of the next instruction in a special register ra (return address)
    jumpal label

  • Jump to address in \(register_1\)
    jumpr r1

  • Copy the value of \(register_2\) to \(register_1\)
    move r1 r2

Code Generation for a Function Call

Code Generation for a Function Definition

• The callee saves the old return address, evaluates its body, pops the return address, pops the args, and then restores the fram pointer

Calling Sequence: Example for \(f(x,y)\)

Calling Sequence: Example for \(f(x,y)\)

Calling Sequence: Example for \(f(x,y)\)

Calling Sequence: Example for \(f(x,y)\)

Code Generation for Variables/Parameters

• Variable references are the last construct

• The “variables” of a function are its parameters:

  • They are in the activation record

  • 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/Parameters

• Solution: use the frame pointer

  • Always points to the return address on the stack

  • Since it does not move, it can be used to find the variables

• Let \(x_i\) be the \(i^{th}\) formal parameter of the function for which code is being generated
  

  \(cgen(x_i) =\) lw acc offset(fp) \(\qquad\); offset = 4 * i

Code Generation for Variables/Parameters

• Example: for a function \(f(x,y) \; begin \; e \; end\), the activation and frame pointer are set up as follows (when evaluating \(e\))

  

Activation Record and Code Generation Summary

• The activation record must be designed together with the code generator

• Code generation can be done by recursive traversal of the AST

• Note: production compilers do different things:

  • emphasis is on keeping values in registers

  • intermediate results are laid out in the activation record, not pushed and popped from the stack

  • as a result, code generation is often performed in synergy with register allocation