# Carnegie Mellon Lecture 6 Register Allocation I. Introduction II. Abstraction and the Problem III. Algorithm Reading: Chapter 8.8.4 Before next class:

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Carnegie Mellon Lecture 6 Register Allocation I. Introduction II. Abstraction and the Problem III. Algorithm Reading: Chapter 8.8.4 Before next class: Chapter 10.1 - 10.2 CS243: Register Allocation1

Carnegie Mellon Pseudo Registers CS243: Register Allocation2 int g; int foo(int a, int b) { if (a+b) { g++; bar(); g--; } return a+b+g; } preg_a = reg2; preg_b = reg3; preg_0 = &g; preg_1 = preg_a + preg_b; preg_g = *preg_0; if(preg_1 == 0) goto Lab; preg_2 = preg_g + 1; *preg_0 = preg_2; bar(); preg_3 = *preg_0; preg_g= preg_3 - 1; *preg_0 = preg_g; Lab :; reg2 = preg_1 + preg_g; return ; Code Selection Machine instructions using pseudo registers Pseudo registers Like machine registers, but infinite number No aliasing

Carnegie Mellon I. Motivation Problem – Allocation of variables (pseudo-registers) to hardware registers in a procedure Perhaps the most important optimization – Directly reduces running time (memory access  register access) – Other optimizations heavily interact PRE Scheduling Loop optimizations CS243: Register Allocation3

Carnegie Mellon Goal Find an assignment for all pseudo-registers, if possible. If there are not enough registers in the machine, choose registers to spill to memory CS243: Register Allocation4

Carnegie Mellon Example CS243: Register Allocation5 B = … = A D = = B + D L1: C = … = A D = = C + D A = … IF A goto L1

Carnegie Mellon II. An Abstraction for Allocation & Assignment Intuitively – Two pseudo-registers interfere if at some point in the program they cannot both occupy the same register. Interference graph: an undirected graph, where – nodes = pseudo-registers – there is an edge between two nodes if their corresponding pseudo-registers interfere What is not represented – Extent of the interference between uses of different variables – Where in the program is the interference CS243: Register Allocation6

Carnegie Mellon Register Allocation and Coloring A graph is n-colorable if: – every node in the graph can be colored with one of the n colors such that two adjacent nodes do not have the same color. Assigning n register (without spilling) = Coloring with n colors – assign a node to a register (color) such that no two adjacent nodes are assigned same registers(colors) Is spilling necessary? = Is the graph n-colorable? To determine if a graph is n-colorable is NP-complete, for n>2 Too expensive Heuristics CS243: Register Allocation7

Carnegie Mellon III. Algorithm Step 1. Build an interference graph a.refining notion of a node b.finding the edges Step 2. Coloring – use heuristics to try to find an n-coloring Success: – colorable and we have an assignment Failure: – graph not colorable, or – graph is colorable, but it is too expensive to color CS243: Register Allocation8

Carnegie Mellon Step 1a. Nodes in an Interference Graph CS243: Register Allocation9 B = … = A D = = B + D L1: C = … = A D = = D + C A = … IF A goto L1 A = 2 = A

Carnegie Mellon Live Ranges and Merged Live Ranges Motivation: to create an interference graph that is easier to color – Eliminate interference in a variable’s “dead” zones. – Increase flexibility in allocation: can allocate same variable to different registers A live range consists of a definition and all the points in a program (e.g. end of an instruction) in which that definition is live. – How to compute a live range? Two overlapping live ranges for the same variable must be merged CS243: Register Allocation10 a = … … = a

Carnegie Mellon Example (Revisited) CS243: Register Allocation11 B = … = A D = (D 2 ) = B + D L1: C = … = A D = (D 1 ) = D + C A = … (A 1 ) IF A goto L1 A = (A 2 ) = D = A {A}{A 1 } {A,C}{A 1,C} {C}{A 1,C} {C,D}{A 1,C,D 1 } {D}{A 1,C,D 1 } {} {A}{A 1 } {A,B}{A 1,B} {B}{A 1,B} {B,D}{A 1,B,D 2 } {D} {A 1,B,D 2 } {D}{A 1,B,C,D 1,D 2 } {A,D}{A 2,B,C,D 1,D 2 } {A}{A 2,B,C,D 1,D 2 } {}{A 2,B,C,D 1,D 2 } (Does not use A, B, C, or D.) liveness reaching-def

Carnegie Mellon Merging Live Ranges Merging definitions into equivalence classes – Start by putting each definition in a different equivalence class – For each point in a program: if (i) variable is live, and (ii) there are multiple reaching definitions for the variable, then: – merge the equivalence classes of all such definitions into one equivalence class From now on, refer to merged live ranges simply as live ranges CS243: Register Allocation12

Carnegie Mellon Step 1b. Edges of Interference Graph Intuitively: – Two live ranges (necessarily of different variables) may interfere if they overlap at some point in the program. – Algorithm: At each point in the program: – enter an edge for every pair of live ranges at that point. An optimized definition & algorithm for edges: – Algorithm: check for interference only at the start of each live range – Faster – Better quality CS243: Register Allocation13

Carnegie Mellon Example 2 CS243: Register Allocation14 A = …L1: B = … IF Q goto L1 IF Q goto L2 L2: … = B … = A

Carnegie Mellon Step 2. Coloring Reminder: coloring for n > 2 is NP-complete Observations: – a node with degree < n  can always color it successfully, given its neighbors’ colors – a node with degree = n  – a node with degree > n  CS243: Register Allocation15

Carnegie Mellon Coloring Algorithm Algorithm: – Iterate until stuck or done Pick any node with degree < n Remove the node and its edges from the graph – If done (no nodes left) reverse process and add colors Example (n = 3): Note: degree of a node may drop in iteration Avoids making arbitrary decisions that make coloring fail CS243: Register Allocation16 B CEA D

Carnegie Mellon What Does Coloring Accomplish? Done: – colorable, also obtained an assignment Stuck: – colorable or not? CS243: Register Allocation17 B CEA D

Carnegie Mellon What if Coloring Fails? Use heuristics to improve its chance of success and to spill code Build interference graph Iterative until there are no nodes left If there exists a node v with less than n neighbor place v on stack to register allocate else v = node chosen by heuristics (least frequently executed, has many neighbors) place v on stack to register allocate (mark as spilled) remove v and its edges from graph While stack is not empty Remove v from stack Reinsert v and its edges into the graph Assign v a color that differs from all its neighbors (guaranteed to be possible for nodes not marked as spilled) CS243: Register Allocation18

Carnegie Mellon Summary Problems: – Given n registers in a machine, is spilling avoided? – Find an assignment for all pseudo-registers, whenever possible. Solution: – Abstraction: an interference graph nodes: live ranges edges: presence of live range at time of definition – Register Allocation and Assignment problems equivalent to n-colorability of interference graph  NP-complete – Heuristics to find an assignment for n colors successful: colorable, and finds assignment not successful: colorability unknown & no assignment CS243: Register Allocation19

Carnegie Mellon Extra Scope of Problem: – Example Showed Whole Function – Separate Inner Loops Interaction with scheduling Don’t want to spill in the inner loop – Inter-procedural Preferencing – Avoid copies on function calls, entries and returns – Glue together inner loop allocation with global Rematerialization CS243: Register Allocation20

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