Lecture 8 AST Visiting and Code Generation

1. Lecture 8 AST Visiting and Code Generation Xiaoyin Wang CS 5363 Programming Languages and Compilers

2. Code Generation • Tree Walking • Code Generation Rules • Control Flow Graph

3. Code Generation • After we constructed the AST and decorated it • We finally start working on generating target code • Here we output 3-address code • Still an intermediate representation • Assuming unlimited registers • No machine specifications • Format: op src1 src2 => dest

4. Code Generation • It is surprisingly simple: • You just need to walk the tree from top to bottom • Apply rules for each AST Node • Then, we are done!

5. Code Generation • Basic Operations • Emit: write an instruction or a label to the instruction stream • Register.Next(): get the next available register • Label.Next(): get the next available label • For expression AST Nodes, add a register field to save the register that stores its value

6. Tree walking • Depth-first tree walking • Always visit children left to right

7. Applying Rules: Binary Operation • Decide what should the code do (semantics) for each AST Node • For example: Consider Add (left, right) • What you want to do is:eval <left> eval <right> add left, right => result

8. Applying Rules: Binary Operation • So in your code: public void visit (Add add){ add.left.accept(this); Register r1 = add.left.reg; add.right.accept(this); Register r2 = add.right.reg; Register r3 = Register.next(); emit(new instruction(‘add’, r3, r1, r2)); add.reg = r3; }

9. Applying Rules: Binary Operation • Another example: • expr1 > expr2 eval <expr1> eval <expr2> set1g expr1, expr2 => result

10. Applying Rules: Binary Operation • Code Example public void visit (Compare comp){ comp.left.accept(this); Register r1 = comp.left.reg; comp.right.accept(this); Register r2 = comp.right.reg; Register r3 = Register.next(); swtich(comp.op){ case ‘>’: emit(new instruction(‘set1g’, r3, r1, r2));break; case ‘>=’: … … } comp.reg = r3; }

11. Applying Rules: Num Literal • For an immediate number • We just need to assign it to a register, which can be used later • Code: public void visit(NumLiteral num){ Register r1 = Register.next(); num.reg = r1; emit( new Instruction (‘mov’, r1, num.value) ); }

12. Applying Rules: Identifier • What we should do: • Just find out which register the value is saved in • If no register is found: error, the variable is not initialized public void visit(Identifier ident){ Register r1 = symbolTable.get(ident.name).reg; if (r1 != null) { ident.reg = r1;} else {throw exception … } }

13. Applying Rules: Assignment • Ident := expr • What we should do • First eval expr • Then find the register of the ident / Or create a new register for the ident • Insert the mov instruction

14. Applying Rules: Assignment • Ident := expr • What we should do • First eval expr • Then find the register of the ident / Or create a new register for the ident • Insert the mov instruction expr.accept(this); Register r1 = expr.reg; r2 = symbolTable.get(ident.name).reg; if(r2 == null){ r2 = Register.next(); symbolTable.get(ident.name).reg = r2; } emit(new Instruction('mov', r2, r1));

15. Applying Rules: If statement • Another example: if expr then stmt1 else stmt2; IfElseStmt thenStmt elseStmt cond

16. Applying Rules: If statement • What we want is: 1. evaluate cond expression 2. if return true: goto then stmt 3. else: goto else stmt 4. goto the end of the stmt on each of the branches

17. Applying Rules: If statement Eval Cond • Control Flow: Code pattern: <eval condition> JZ reg ELSE <then statements> JMP END ELSE: <else statements> END: True? JZ thenStmt elseStmt JMP end

18. Applying Rules: If statement • So in your code: public Register visit (IfElseStmtifelse){ ifelse.cond.accept(this); Register r1 = ifelse.cond.reg; Label elseLabel = Label.next(); Label endLabel = Label.next(); emit (new Instruction(‘JZ’, r1, elseLabel)) ifelse.thenStmt.accept(this); emit (new Instruction(‘JMP’, endLabel)) emit (elseLabel); ifelse.elseStmt.accept(this); emit (endLabel) }

19. Applying Rules: while statement while cond do stmt end WhileStmt stmt cond

20. Applying Rules: while statement • What we want is: 1. Set a label Start before evaluate cond expression 2. evaluate cond expression 3. if false: goto label End 4. else: go on to execute stmt 5. goto label Start

21. Applying Rules: while statement • Control Flow: Code pattern: Eval Cond Start: <eval condition> JZ reg END <stmt> JMP Start END: True? JMP JZ stmt end

22. Applying Rules: while statement • So in your code: public Register visit (WhileStmt while){ Label startLabel = Label.next(); emit(startLabel) while.cond.accept(this); Register r1 = while.cond.reg; Label endLabel = Label.next(); emit (new Instruction(‘JZ’, r1, endLabel)) while.stmt.accept(this); emit (new Instruction(‘JMP’, startLabel)) emit (endLabel) }

23. Instruction Set The Appendix A of MIPS Assembly Specification On the course website A more brief summary at http://xywang.100871.net/SPIM_instr.pdf

24. Instruction Set Registers and Memory address Memory instructions Arithmetic instructions Comparison instructions Branch instructions System call related instructions

25. Registers and Memory address (A-24) $zero: 0 $t0 – $t9: local variables $s0 – $s7: global variables $a0 – $a3: function arguments $v0, v1: function return values $fp: frame pointer (first byte) $sp: stack pointer (last byte)

26. Memory Instructions (A65) li: Load immediate value Example: li $t0, 2 la: Load address Example: la $t0, str lw: Load value from memory Example: lw $t0, -4($fp) sw: save value to memory Example: sw $t0, -4($fp)

27. Arithmetic instructions (A51-A57) add / addu / addi Example: addu $t0, $t1, $t2 sub / subu mul / mulo / mulou div / divu rem / remu

28. Comparison instructions (A57-A59) slt : set 1 if less than Example: slt $t0, $t1, $t2 slti: set 1 if less than immediate Example: slt $t0, $t1, 3 seq: set 1 if equal sgt: set 1 if greater sge, sle, sne

29. Branch instructions (A59-A63) j/b: jump to label Example: j bbb beqz / bnez: jump to label if equal or not equal to zero Example: beqz $t0 bbb beq, bne, bgt, …

30. System calls: for read/write int System call code: in $v0 1: print int, parameter in $a0 5: read int, result in $v0 Example: li $v0, 4 la $a0, newline syscall li $v0, 5 syscall add $t0, $v0, $zero

31. Control Flow Graph • An intermediate presentation of target code • Enables flow-oriented code optimization • Better understanding of the code

32. Basic Blocks • A block of the code that will be sequentially executed • With no jump instructions and labels for other instructions to jump to • Nodes in the control flow graph

33. Generate Control Flow Graph with Basic Blocks • In the generated code • Step 1: split the code to a number of basic blocks • Step 2: find the relationship between all basic blocks, for each basic block: • Its predecessors are the basic block directly before it (if no unconditional jump at the end), and all blocks can jump to it • Its successors are the basic block directly after it (if no unconditional jump at the end), and all blocks it jumps to

34. mov 0 => r_SQRT readint => r_N mov 0 => r_SQRT readint => r_N L1: cmp_LE r1, r_N => r2 jumpeqz r2 -> L2 add r_SQRT, 1 => r4 mov r4 => r_SQRT jump -> L1 L2: sub r_SQRT, r5 => r6 writeInt r_SQRT exit L1: cmp_LE r1, r_N => r2 jumpeqz r2 -> L2 break add r_SQRT, 1 => r4 mov r4 => r_SQRT jump -> L1 L2: sub r_SQRT, r5 => r6 writeInt r_SQRT exit

35. Block 1: mov 0 => r_SQRT readint => r_N Block 2: L1: cmp_LE r1, r_N => r2 jumpeqz r2 -> L2 Block 3: add r_SQRT, 1 => r4 mov r4 => r_SQRT jump -> L1 Block 4: L2: sub r_SQRT, r5 => r6 writeInt r_SQRT exit

36. Basic Transformation From 3 address code to MIPS code 3 address code: Assumes unlimited number of registers MIPS code: Only limited number of registers, but memory can be normally viewed as unlimited

37. Basic Transformation From 3 address code to MIPS code Therefore Any: Op r_x, r_y => r_z Can transform to: lw $t1, -4*x($fp) lw $t2, -4*y($fp) Op $t0, $t1, $t2 sw $t0, -4*z($fp)