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1 Instruction Selection DAG Tiling Problems Pentium ISA Example Tiling Instruction Selection CS412/CS413 1. Translate low-level IR code into DAG representation Introduction to Compilers 2. Then find a good tiling of the DAG Tim Teitelbaum - disjoint set of tiles that cover the DAG - Maximal munch algorithm Lecture 32: More Instruction Selection - Dynamic programming algorithm 15 Apr 07 CS 412/413 Spring 2007 Introduction to Compilers 1 CS 412/413 Spring 2007 Introduction to Compilers 2 DAG Tiling Problems • Classes of registers •Goal:find a good covering of DAG with tiles – Registers may have specific purposes – Example: Pentium multiply instruction •Problem:need to know what variables are in registers - multiply register eax by contents of another register - store result in eax (low 32 bits) and edx (high 32 bits) • Assume abstract assembly: - need extra instructions to move values into eax – Machine with infinite number of registers • Two-address machine instructions – Temporary/local variables stored in registers – Three-address low-level code – Parameters/heap variables: use memory accesses – Need multiple machine instructions for a single tile •CISC versus RISC – Complex instruction sets => many possible tiles and tilings – Example: multiple addressing modes (CISC) versus load/store architectures (RISC) CS 412/413 Spring 2007 Introduction to Compilers 3 CS 412/413 Spring 2007 Introduction to Compilers 4 Pentium ISA Example Tiling •Consider: t = t + i •Pentium:two-address CISC architecture t = temporary variable i = parameter • Multiple addressing modes: source operands may be = – Immediate value: imm • Need new temporary registers t + – Register: reg between tiles (unless operand t1 t – Indirect address: [reg], [imm], [reg+imm], node is labeled with temporary) load {i} – Indexed address: [reg+reg’], [reg+imm*reg’], t0 [reg+imm*reg’+imm’] • Resulting code: - mov %ebp, t0 ebp 20 • Destination operands = same, except immediate values sub $20, t0 mov 0(t0), t1 add t1, t CS 412/413 Spring 2007 Introduction to Compilers 5 CS 412/413 Spring 2007 Introduction to Compilers 6 1 Tiles Some Tiles t2 store + mov t1, t2 = mov t1 mov t2, t1 + add $1, t2 t1 t2 $10, 0(t1,t2) 10 1 t1 t2 • Tiles capture compiler’s understanding of instruction set t3 • Each tile: sequence of machine instructions that mov t2, t3 mov t1, %eax t3 match a subgraph of the DAG + add t1, t3 mul t2 * • May need additional move instructions t1 t2 mov %eax, t3 t1 t2 • Tiling = cover the DAG with tiles CS 412/413 Spring 2007 Introduction to Compilers 7 CS 412/413 Spring 2007 Introduction to Compilers 8 Conditional Branches Maximal Munch Algorithm • How to tile a conditional jump? • Maximal Munch = find largest tiles (greedy algorithm) • Fold comparison into tile • Start from top of tree • Find largest tile that matches top node • Tile remaining subtrees recursively tjump tjump store t1 L == L + 4 t1 t2 load * test t1,t1 cmp t1,t2 + 4 load jnz L je L ebp 8 + ebp 12 CS 412/413 Spring 2007 Introduction to Compilers 9 CS 412/413 Spring 2007 Introduction to Compilers 10 DAG Representation Example •DAG:a node may have multiple parents • Algorithm: same, but a node with multiple parents x = x + 1; occurs inside a tile only if all its parents are in the tile store store + + + load * ebp 8 load 1 + 4 load + ebp 8 + ebp 8 ebp 12 CS 412/413 Spring 2007 Introduction to Compilers 11 CS 412/413 Spring 2007 Introduction to Compilers 12 2 Example Alternate (CISC) Tiling x = x + 1; x = x + 1; store store add $1, 8(%ebp) t2 mov 8(%ebp), t1 + + + + t1 mov t1, t2 = ebp 8 load 1 add $1, t2 ebp 8 load 1 + + mov t2, 8(%ebp) + r/m32 const ebp 8 ebp 8 r/m32 CS 412/413 Spring 2007 Introduction to Compilers 13 CS 412/413 Spring 2007 Introduction to Compilers 14 ADD Expression Tiles ADD Statement Tiles = t1 t1 Intel Architecture + mov t2, t1 r/m32 const + + add r/m32, t1 t2 r/m32 add imm32, %eax t2 t3 add imm32, r/m32 = add imm8, r/m32 + t1 add r32, r/m32 r/m32 mov t2, t1 + add r/m32, r32 = add imm32, t1 t2 const + r32 r/m32 CS 412/413 Spring 2007 Introduction to Compilers 15 CS 412/413 Spring 2007 Introduction to Compilers 16 Designing Tiles More Handy Tiles • Only add tiles that are useful to compiler lea instruction computes a memory address • Many instructions will be too hard to use effectively or will offer no advantage t3 • Need tiles for all single-node trees to guarantee lea (t1,t2), t3 + that every tree can be tiled, e.g. t1 t2 t3 t1 mov t2, t1 + + add t3, t1 lea c1(t1,t2,c2), t3 + * t2 t3 t1 c1 c2 t2 CS 412/413 Spring 2007 Introduction to Compilers 17 CS 412/413 Spring 2007 Introduction to Compilers 18 3 Matching Jump for RISC Condition Code ISA • As defined in lecture, have • Pentium: condition encoded in jump instruction tjump(cond, destination) • cmp: compare operands and set flags fjump(cond, destination) • jcc: conditional jump according to flags • Our tjump/fjump translates easily to RISC ISAs that have explicit comparison result set condition codes tjump cmp t1, t2 tjump MIPS < L jl L t1 cmplt t2, t3, t1 t2 < L t3 br t1, L test condition codes t2 t3 CS 412/413 Spring 2007 Introduction to Compilers 19 CS 412/413 Spring 2007 Introduction to Compilers 20 Fixed-register instructions Implementation mul r/m32 • Maximal Munch: start from top node Multiply value in register eax • Find largest tile matching top node and all of the Result: low 32 bits in eax, high 32 bits in edx children nodes jecxz L • Invoke recursively on all children of tile Jump to label L if ecx is zero • Generate code for this tile • Code for children will have been generated already in add r/m32, %eax recursive calls Add to eax • How to find matching tiles? • No fixed registers in low IR except frame pointer • Need extra move instructions CS 412/413 Spring 2007 Introduction to Compilers 21 CS 412/413 Spring 2007 Introduction to Compilers 22 Matching Tiles Tile Specifications abstract class LIR_Stmt { Assembly munch(); • Previous approach simple, efficient, but hard-codes tiles } and their priorities class LIR_Assign extends LIR_Stmt { LIR_Expr src, dst; • Another option: explicitly create data structures Assembly munch() { representing each tile in instruction set if (src instanceof IR_Plus && = ((IR_Plus)src).lhs.equals(dst) && – Tiling performed by a generic tree-matching and + is_regmem32(dst)) { code generation procedure Assembly e = ((LIR_Plus)src).rhs.munch(); – Can generate from instruction set description: return e.append(new AddIns(dst, r/m32 e.target())); code generator generators } – For RISC instruction sets, over-engineering else if ... } } CS 412/413 Spring 2007 Introduction to Compilers 23 CS 412/413 Spring 2007 Introduction to Compilers 24 4 How Good Is It? Improving Instruction Selection • Very rough approximation on modern pipelined • Because it is greedy, Maximal Munch does not architectures: execution time is number of tiles necessarily generate best code – Always selects largest tile, but not necessarily the • Maximal munch finds a locally optimal (two adjacent fastest instruction tiles can never be combined into one) but not – May pull nodes up into tiles inappropriately – it may necessarily globally optimum tiling (least cost of all covers) be better to leave below (use smaller tiles above and larger, or faster tiles below) • Metric used: tile size • Better to use dynamic programming, an optimization technique that uses memoization to assure that subproblems are never solved more than once. CS 412/413 Spring 2007 Introduction to Compilers 25 CS 412/413 Spring 2007 Introduction to Compilers 26 Timing Cost Model Finding globally optimum tiling •Idea:associate cost with each tile (say proportional to •Goal:find minimum total cost tiling of DAG number of cycles to execute) – may not be a good metric on modern architectures • Algorithm: for every node, find minimum total cost tiling • Total execution time is sum of costs of all tiles of that node and subgraph below it Cost = 2 store • Lemma: Given minimum cost tiling of all nodes in + + Cost=1 subgraph, we can find minimum cost tiling of the node by trying out all possible tiles matching the node Total cost: 5 ebp 8 load 1 • Therefore: start from leaves, work upward to top node + ebp 8 Cost = 2 CS 412/413 Spring 2007 Introduction to Compilers 27 CS 412/413 Spring 2007 Introduction to Compilers 28 Dynamic Programming: a[i] Recursive Implementation mov 8(%ebp), t1 • Traverse DAG recursively, and for each node n, record <t,c>, where load mov 12(%ebp), t2 mov (t1,t2,4), t3 – t is the best tile to use for subgraph rooted at n, – c is the total cost of tiling the subgraph rooted at n if t is chosen. + • To compute <t,c> for node n load * – Consider every tile t’ that matches rooted at n, and compute 4 load total cost c’ = cost of tile t’ + sum of the costs of tiling the + subgraphs rooted at the leaves of t’ (which costs can be + computed recursively and memoized) ebp 8 ebp 12 – Store lowest-cost tile t’ and its total cost c’ • To emit code, traverse least-cost tiles recursively and emit code in postorder CS 412/413 Spring 2007 Introduction to Compilers 29 CS 412/413 Spring 2007 Introduction to Compilers 30 5 Memoization Problems with Model class IR_Move extends IR_Stmt { = • Modern processors: IR_Expr src, dst; + – execution time not sum of tile times Assembly best; // initialized to null – instruction order matters int optTileCost() { if (best != null) return best.cost(); r/m32 • Processors pipeline instructions and execute if (src instanceof IR_Plus && different pieces of instructions in parallel ((IR_Plus)src).lhs.equals(dst) && is_regmem32(dst)) { • bad ordering (e.g.
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