1. 1 Copy Propagation What does it mean? Given an assignment x = y, replace later uses of x with uses of y, provided there are no intervening assignments to x or y. Similar to register coalescing, which eliminates copies from one register to another. When is it performed? At any level, but usually early in the optimization process. What is the result? Smaller code

2. 2 Local Copy Propagation Local copy propagation Performed within basic blocks Algorithm sketch: traverse BB from top to bottom maintain table of copies encountered so far modify applicable instructions as you go

3. 3 Local Copy Propagation Algorithm sketch for a basic block containing instructions i 1, i 2,..., i n for instr = i 1 to i n if instr is of the form 'res = opd1 op opd2' opd1 = REPLACE(opd1, copytable) opd2 = REPLACE(opd2, copytable) else if instr is of the form 'res = var' var = REPLACE(var, copytable) if instr has a lhs res, REMOVE from copytable all pairs involving res. if instr is of the form 'res = var' /* i.e. a copy */ insert {(res, var2)} in the copytable endfor REPLACE(opd, copytable) if you find (opd, x) in copytable /* use hashing for faster access */ return x else return opd copytable is table containing copy pairs e.g. if there's an assignment x := a, then copytable should contain (x,a)

4. 4 Local Copy Propagation step instruction updated instruction table contents 1 b = a b = a{(b,a)} 2 c = b + 1 c = a + 1{(b,a)} 3d = b d = a{(b,a), (d,a)} 4 b = d + c b = a + c{(d,a)} 5 b = d b = a{(d,a), (b,a)} Example: Local copy propagation on basic block: b = a c = b + 1 d = b b = d + c b = d Note: if there was a definition of 'a' between 3 and 4, then we would have to remove (b,a) and (d,a) from the table. As a result, we wouldn't be able to perform local copy propagation at instructions 4 and 5. However, this will be taken care of when we perform global copy propagation.

5. 5 Copy Propagation Global copy propagation Performed on flow graph. Given copy statement x=y and use w=x, we can replace w=x with w=y only if the following conditions are met: 1. x=y must be the only definition of x reaching w=x This can be determined through ud-chains 2. There may be no definitions of y on any path from x=y to w=x. Use iterative data flow analysis to solve this. Even, better, use iterative data flow analysis to solve both problems at the same time. x=y w=x

6. 6 Copy Propagation Data flow analysis to determine which instructions are candidates for global copy propagation forward direction gen[Bi] = {(x,y,i,p) | p is the position of x=y in block Bi and neither x nor y is assigned a value after p} kill[Bi] = {(x,y,j,p) | x=y, located at position p in block Bj  Bi, is killed due to a definition of x or y in Bi } in[B]=  out[P] over the predecessors P of B Initialize in[B1]= , in[B]=U for B  B1 p: x = y... generate x=y if no definitions of x or y in this area... q: x = z... s: y = w kill all other definitions of x kill all other definitions of y

7. 7 Copy Propagation entry exit 1: c = a + b 2: d = c 3: e = d * d 4: f = a + c 5: g = e 6: a = g + d 7: a < c 8: h = g + 1 9: f = d - g 10: f > a 11: b = g * a 12: h < f entry exit c = a + b d = c e = c * c f = a + c g = e a = e + c a < c h = e + 1 f = c - e f > a b = e * a h < f dead code?

8. 8 Copy Propagation Copy propagation will not detect the opportunity to replace x with y in the last block below: Copy propagation may generate code that does not need to be evaluated any longer. This will be handled by optimizations that perform redundancy elimination. z >0 x = y w = x + z Mini quiz: which optimization can handle this? Answer: If we perform an optimization similar to code hoisting (i.e. one that would move the copy either up or down the graph) then copy propagation will be able to update "w=y+z"

9. 9 Constant Propagation What does it mean? Given an assignment x = c, where c is a constant, replace later uses of x with uses of c, provided there are no intervening assignments to x. Similar to copy propagation Extra feature: It can analyze constant-value conditionals to determine whether a branch should be executed or not. When is it performed? Early in the optimization process. What is the result? Smaller code Fewer registers

10. 10 Common Subexpression Elimination Local common subexpression elimination Performed within basic blocks Algorithm sketch: Traverse BB from top to bottom Maintain table of expressions evaluated so far if any operand of the expression is redefined, remove it from the table Modify applicable instructions as you go generate temporary variable, store the expression in it and use the variable next time the expression is encountered. x = a + b... y = a + b t = a + b x = t... y = t

11. 11 Common Subexpression Elimination c = a + b d = m * n e = b + d f = a + b g = - b h = b + a a = j + a k = m * n j = b + d a = - b if m * n go to L t1 = a + b c = t1 t2 = m * n d = t2 t3 = b + d e = t3 f = t1 g = -b h = t1 /* commutative */ a = j + a k = t2 j = t3 a = -b if t2 go to L the table contains quintuples: (pos, opd1, opr, opd2, tmp)

12. 12 Common Subexpression Elimination Global common subexpression elimination Performed on flow graph Requires available expression information In addition to finding what expressions are available at the endpoints of basic blocks, we need to know where each of those expressions was most recently evaluated (which block and which position within that block).

13. 13 Common Subexpression Elimination Global common subexpression elimination Algorithm sketch: For each block B and each statement x=y+z, s.t. {y+z}  in[B] i. Find the evaluation of y+z that reaches B, say w=x+y ii. Create temporary variable t iii. Replace [w=y+z] with [t=y+z; w=t] iv. Replace [x=y+z] with [x=t] Notes: This method will miss the fact that b and d have the same value: a = x+y c = x+y b = a * z d = c * z Answer: Value Numbering Mini quiz: which optimization can handle this?

14. 14 Common Subexpression Elimination entry exit c = a + b d = a * c e = d * d f = a + b c = c * 2 c > d g = a * cg = d * d g > 10 entry exit t1 = a + b c = t1 d = a * c t2 = d * d e = t2 f = t1 c = c * 2 c > d g = a * cg = t2 g > 10

15. 15 Loop Optimizations Important: Loop bodies execute frequently. We will examine: Loop-invariant code motion Identify computations that have the same value at every iteration and move them outside the loop, so they are performed only once. Not always a good idea: Induction variable elimination Induction variables are incremented/decremented by a constant amount at each iteration, typically related to the loop control variable. If we can identify them and take advantage of this, we can reduce the number of calculations as well as certain data dependences. cin >> n; for (int i=0; i<n; i++) x = y/z;

16. 16 Loop-Invariant Code Motion Step 1: Identify loop-invariant code. An instruction is loop-invariant if, for each operand: 1. The operand is constant, OR 2. All definitions of that operand that reach the instruction are outside the loop, OR 3. There is exactly one in-loop definition of the operand that reaches the instruction, and that definition is loop invariant. Is this necessary? How do we determine #2? Use ud-chains

17. 17 Loop-Invariant Code Motion Step 2: Move it outside the loop. Algorithm sketch: – For each loop-invariant instruction i: x=y+z in basic block B, check that it satisfies the conditions: i. B dominates all exits of the loop ii. x is not defined anywhere else in the loop iii. B dominates all uses of x in the loop (i.e. all uses of x in the loop can be reached only by i – Move each instruction i that satisfies these requirements to a newly created pre-header of the loop, making certain that any non- constant operands (such as y, z) have already had their definitions moved to the pre-header. Note: When applying loop-invariant code motion to nested loops, work from the innermost loop outwards.

18. 18 Loop-Invariant Code Motion Why are these conditions necessary? i. B dominates all exits of the loop i = 1 x = 5 y = a+b y > 0 x = 2 y-- i++ i<n z = x x=2 is loop invariant, but does not dominate the exit. If we move it outside the loop, we will get different results in the cases when control flows along the green line.

19. 19 Loop-Invariant Code Motion Why are these conditions necessary? ii. x is not defined anywhere else in the loop i = 1 x = 5 y = a+b x = 3 y > 0 x = 2 y-- i++ i<n z = x x=3 is loop invariant and dominates the exit. However, if we move it outside the loop, we will get different results in the cases when control flows along the green lines.

20. 20 Loop-Invariant Code Motion Why are these conditions necessary? iii. B dominates all uses of x in the loop (i.e. all uses of x in the loop can be reached only by i i = 1 x = 5 y = x+b x = 3 y > 0 print y y-- i++ i<n z = x x=3 is loop invariant, it dominates the exit and x is not defined anywhere else in the loop. However, if we move it outside the loop, we will get a different value for y at the first iteration, when x should be 5

21. 21 Loop-Invariant Code Motion entry exit b = 2 i = 1 d = a + d e = 1 + d d = -c f = 1 + a i = i+1 a < 2 a = b+1 c = 2 i mod 2 = 0 Exercise: What happens if you perform constant propagation followed by constant folding after the loop-invariant code motion in this loop? T F Reminder: An instruction is loop-invariant if, for each operand: 1. The operand is constant, OR 2. All definitions of that operand that reach the instruction are outside the loop, OR 3. There is exactly one in-loop definition of the operand that reaches the instruction, and that definition is loop invariant. Reminder: Move a loop-invariant x=... located in B, if 1. B dominates all exits of the loop 2. x is not defined anywhere else in the loop 3. B dominates all uses of x in the loop

22. 22 Loop-Invariant Code Motion entry exit b = 2 i = 1 d = a + d e = 1 + d d = -c f = 1 + a i = i+1 a < 2 a = b+1 c = 2 i mod 2 = 0 entry exit b = 2 i = 1 a = b+1 c = 2 t1 = a<2 d = a + d e = 1 + d d = -c f = 1 + a i = i+1 t1 i mod 2 = 0 T F T F

23. 23 after constant propagation and constant folding entry exit b = 2 i = 1 a = b+1 c = 2 t1 = a<2 d = a + d e = 1 + d d = -c f = 1 + a i = i+1 t1 i mod 2 = 0 T F entry exit b = 2 i = 1 a = 3 c = 2 t1 = false d = a + d e = 1 + d d = -c f = 1 + a i = i+1 t1 i mod 2 = 0 F

24. 24 Induction variable elimination We will concentrate on two kinds of induction variables: Basic These are of the form i = i  c, where c is a constant How about i=i*c or i=i/c ? Derived These are defined only once in the loop, and their value is a linear function of a basic induction variable. E.g. j = c 1 *i+c 2. j is expressed as (i, c 1, c 2 ) and is said to belong to the family of i What information do we need? Reaching definitions Loop-invariants

25. 25 Induction variable elimination Algorithm sketch: Find all basic induction variables Find all assignments of the form k=c op j, where op  {+, -, *, /} and j is an induction variable If j is basic, then k belongs to its family. Determine its triple. If j is not basic, then it belongs to the family of a basic induction variable i. Check that there is no assignment to i between j= and k= Check that j= dominates k= Determine k's triple. e.g. if j is (i, a, b) and k=c*j+d, then k's triple is (i, a*c, b*c+d) Perform strength reduction on induction variables

26. 26 Induction variable elimination Performing strength reduction on induction variables Consider: i=i+1 // (i, 1, 1) j=4*i // (i, 4, 4) Goal: i=i+1 j=4*i // initially j=j+4 // afterwards How? Create a new variable s Set s=4*i in the pre-header Set s=s+4 every time i is updated Replace j=4*i with j=s.