Compilation 0368-3133 (Semester A, 2013/14)

1. Compilation 0368-3133 (Semester A, 2013/14) Lecture 8: Activation Records + Register Allocation Noam Rinetzky Slides credit: Roman Manevich, MoolySagiv and EranYahav

2. Registers • Most machines have a set of registers, dedicated memory locations that • can be accessed quickly, • can have computations performed on them, and • Usages • Operands of instructions • Store temporary results • Can (should) be used as loop indexes due to frequent arithmetic operation • Used to manage administrative info • e.g., runtime stack

3. Registers • Number of registers is limited • Need to allocate them in a clever way • Using registers intelligently is a critical step in any compiler • A good register allocator can generate code orders of magnitude better than a bad register allocator

4. simple code generation • assume machine instructions of the form • LD reg, mem • ST mem, reg • OP reg,reg,reg • assume that we have all registers available for our use • Ignore registers allocated for stack management • Treat all registers as general-purpose

5. “Abstract” Code • Fixed number of Registers • Instructions • Load: Memory  Register • Store: Register  Memory • Operation:R1 = R2 + R3 (*) • Assume all registers are available • Ignore registers allocated for stack management • Treat all registers as general-purpose

6. Simple approach • Straightforward solution: • Allocate each variable in activation record • At each instruction, bring values needed into registers, perform operation, then store result to memory mov 16(%ebp), %eaxmov 20(%ebp), %ebxadd %ebx, %eaxmov %eax, 24(%ebx) x = y + z Problem: program execution very inefficient–moving data back and forth between memory and registers

7. Registers in Compound Expressions Why Naïve? • Use registers to store temporaries • Why can we do it? • Maintain a counter for temporaries in c • Initially: c = 0 • cgen(e1op e2) = { Let A = cgen(e1)c = c + 1 Let B = cgen(e2)c = c + 1 Emit( _tc = A op B; ) // _tc is a registerReturn _tc}

8. Register allocation • In TAC, there are an unlimited number of variables • On a physical machine there are a small number of registers: • x86 has four general-purpose registers and a number of specialized registers • MIPS has twenty-four general-purpose registers and eight special-purpose registers • Register allocation is the process of assigning variables to registers and managing data transfer in and out of registers

9. Improving cgen for expressions • Observation – naïve translation needlessly generates temporaries for leaf expressions • Observation – temporaries used exactly once • Once a temporary has been read it can be reused for another sub-expression • cgen(e1op e2) = { Let _t1 = cgen(e1) Let _t2 = cgen(e2) Emit( _t =_t1 op _t2; ) Return t} • Temporaries cgen(e1) can be reused in cgen(e2)

10. Sethi-Ullman translation • Algorithm by Ravi Sethi and Jeffrey D. Ullman to emit optimal TAC • Minimizes number of temporaries for a single expression

11. Example: b*b-4*a*c - * * 4 * b b a c

12. Example (simple): b*b-4*a*c 9 - 3 8 * * 1 7 4 2 4 * b b 5 6 a c

13. Example (optimized): b*b-4*a*c 3 - 2 2 * * 1 2 1 1 4 * b b 1 1 a c

14. Spilling • Even an optimal register allocator can require more registers than available • Need to generate code for every correct program • The compiler can save temporary results • Spill registers into temporaries • Load when needed • Many heuristics exist

15. Simple Spilling Method • Heavy tree – Needs more registers than available • A `heavy’ tree contains a `heavy’subtree whose dependents are ‘light’ • Generate code for the light tree • Spill the content into memory and replace subtree by temporary • Generate code for the resultant tree

16. Example (optimized): b*b-4*a*c 3 - 2 2 * * 1 2 1 1 4 * b b 1 1 a c

17. Example (spilled): x := b*b-4*a*c 2 2 - * 1 t7 1 2 1 * b b 2 1 4 * 1 1 a c x := t7 – 4 * a * c t7 := b * b

18. Simple Spilling Method

19. Generalizations • More than two arguments for operators • Function calls • Register/memory operations • Multiple effected registers • Multiplication • Spilling • Need more registers than available

20. Register Memory Operations • Add_Mem X, R1 • Mult_Mem X, R1 • No need for registers to store right operands Hidden Registers

21. Example: b*b-4*a*c 2 - 1 1 Mult_Mem Mult_Mem 0 1 0 1 Mult_Mem 4 b b 0 1 a c

22. Can We do Better? • Yes: Increase view of code • Simultaneously allocate registers for multiple expressions • But: Lose per expression optimality • Works well in practice

23. Basic Blocks • basic blockis a sequence of instructions with • single entry (to first instruction), no jumps to the middle of the block • single exit (last instruction) • code execute as a sequence from first instruction to last instruction without any jumps • edge from one basic block B1 to another block B2 when the last statement of B1 may jump to B2

24. B1 control flow graph prod := 0 i := 1 … B1 B2 B2 t1 := 4 * i t2 := a [ t1 ] t3 := 4 * i t4 := b [ t3 ] t5 := t2* t4 t6:= prod + t5 prod := t6 t7:= i + 1 i := t7 if i <= 20 gotoB2 … • A directed graph G=(V,E) • nodes V = basic blocks • edges E = control flow • (B1,B2)  E when control from B1 flows to B2

25. Another Example B1 t1 := 4 * i t2 := a [ t1 ] if t2 <= 20 gotoB3 … B1 B2 False True B3 t3 := 4 * i t4 := b [ t3 ] gotoB4 B2 B3 t5 := t2* t4 t6:= prod + t5 prod := t6 gotoB4 B4 … B4 t7:= i + 1 i := t2 Goto B5

26. Creating Basic Blocks • Input: A sequence of three-address statements • Output: A list of basic blocks with each three-address statement in exactly one block • Method • Determine the set of leaders(first statement of a block) • The first statement is a leader • Any statement that is the target of a jump is a leader • Any statement that immediately follows a jump is a leader • For each leader, its basic block consists of the leader and all statements up to but not including the next leader or the end of the program

27. CFG IR example for i from 1 to 10 do for j from 1 to 10 do a[i, j] = 0.0; for i from 1 to 10 do a[i, i] = 1.0; source B1 i = 1 B2 j = 1 t1 = 10*I t2 = t1 + j t3 = 8*t2 t4 = t3-88 a[t4] = 0.0 j = j + 1 if j <= 10 gotoB3 B3 B4 i=i+1 if i <= 10 gotoB2 B5 i = 1 B6 t5=i-1 t6=88*t5 a[t6]=1.0 i=i+1 if I <=10 gotoB6 i = 1 j =1 t1 = 10*I t2 = t1 + j t3 = 8*t2 t4 = t3-88 a[t4] = 0.0 j = j + 1 if j <= 10 goto (3) i=i+1 if i <= 10 goto (2) i=1 t5=i-1 t6=88*t5 a[t6]=1.0 i=i+1 if I <=10 goto(13)

28. Example: Code Block { int n; n := a + 1; x := b + n * n + c; n := n + 1; y := d * n; }

29. Example: Basic Block { int n; n := a + 1; x := b + n * n + c; n := n + 1; y := d * n; }

30. AST of the Example { int n; n := a + 1; x := b + n * n + c; n := n + 1; y := d * n; }

31. Optimized Code (gcc) { intn; n := a + 1; x := b + n * n + c; n := n + 1; y := d * n; }

32. Register Allocation for B.B. • Dependency graphs for basic blocks • Transformations on dependency graphs • From dependency graphs into code • Instruction selection • linearizations of dependency graphs • Register allocation • At the basic block level

33. Dependency graphs • TAC imposes an order of execution • But the compiler can reorder assignments as long as the program results are not changed • Define a partial order on assignments • a < b  a must be executed before b • Represented as a directed graph • Nodes are assignments • Edges represent dependency • Acyclic for basic blocks

34. Running Example { int n; n := a + 1; x := b + n * n + c; n := n + 1; y := d * n; }

35. Sources of dependency • Data flow inside expressions • Operator depends on operands • Assignment depends on assigned expressions • Data flow between statements • From assignments to their use • Pointers complicate dependencies

36. Sources of dependency • Order of subexpresion evaluation is immaterial • As long as inside dependencies are respected • The order of uses of a variable X are immaterial as long as: • X is used between dependent assignments • Before next assignment to X

37. Creating Dependency Graph from AST • Nodes AST becomes nodes of the graph • Replaces arcs of AST by dependency arrows • Operator  Operand • Create arcs from assignments to uses • Create arcs between assignments of the same variable • Select output variables (roots) • Remove ; nodes and their arrows

38. Running Example

39. Dependency Graph Simplifications • Short-circuit assignments • Connect variables to assigned expressions • Connect expression to uses • Eliminate nodes not reachable from roots

40. Running Example

41. Cleaned-Up Data Dependency Graph

42. Common Subexpressions • Repeated subexpressions • Examplesx = a * a + 2 * a * b + b * b;y = a * a – 2 * a * b + b * b;n[i] := n[i] +m[i] • Can be eliminated by the compiler • In the case of basic blocks rewrite the DAG

43. From Dependency Graph into Code • Linearize the dependency graph • Instructions must follow dependency • Many solutions exist • Select the one with small runtime cost • Assume infinite number of registers • Symbolic registers • Assign registers later • May need additional spill • Possible Heuristics • Late evaluation • Ladders

44. Pseudo Register Target Code

45. Non optimized vs Optimized Code { intn; n := a + 1; x := b + n * n + c; n := n + 1; y := d * n; }

46. Register Allocation • Maps symbolic registers into physical registers • Reuse registers as much as possible • Graph coloring (next) • Undirected graph • Nodes = Registers (Symbolic and real) • Edges = Interference • May require spilling

47. Register Allocation for Basic Blocks • Heuristics for code generation of basic blocks • Works well in practice • Fits modern machine architecture • Can be extended to perform other tasks • Common subexpression elimination • But basic blocks are small • Can be generalized to a procedure

49. Global Register Allocation

50. Variable Liveness y = 42 z = 73 x = y + z print(x); x undef, y live, z undef x undef, y live, z live x is live, y dead, z dead x is dead, y dead, z dead • A statement x = y + z • defines x • uses y and z • A variable x is live at a program point if its value (at this point) is used at a later point (showing state after the statement)