Midterm 2 Sample solution

Transcription

1 Nme: Instructions Midterm 2 Smple solution CMSC 430 Introduction to Compilers Fll 2012 November 28, 2012 This exm contins 9 pges, including this one. Mke sure you hve ll the pges. Write your nme on the top of this pge before strting the exm. Write your nswers on the exm sheets. If you finish t lest 15 minutes erly, bring your exm to the front when you re finished; otherwise, wit until the end of the exm to turn it in. Plese be s quiet s possible. If you hve question, rise your hnd. If you feel n exm question ssumes something tht is not written, write it down on your exm sheet. Brring some unforeseen error on the exm, however, you shouldn t need to do this t ll, so be creful when mking ssumptions. Question Score Mx Totl 100 1

2 Question 1. Short Answer (20 points).. (6 points) Briefly explin wht bsic block is. Answer: A bsic block is sequence of sttements such tht () there re no jumps from the bsic block except fter its lst sttement nd (b) there re no jumps into the bsic block except to its initil sttement. b. (7 points) Briefly explin wht stte trnsformtion is in dynmic softwre updting nd why it my be needed. Answer: Stte trnsformtion is the process of modifying the ppliction stte to be comptible with new progrm version. Stte trnsformtion is needed becuse often code updtes code with dt representtion nd type chnges. 2

3 c. (7 points) Briefly explin wht time trvel debugger is. Answer: Typicl interctive debuggers llow the developer to set brekpoints nd then, whenever execution is stopped, inspect the progrm stte t tht point. A time-trvel debugger dditionlly llows the developer to inspect pst progrm sttes. (A cnonicl opertion in time-trvel debugger is step bckwrds, which runs execution one step in reverse.) 3

4 Question 2. Subtyping (14 points). Suppose tht int is subtype of flot. To sve writing, let s write i for int nd f for flot.. (7 points) Write down every type t such tht t (i i f), following stndrd subtyping rules. Hint: It my be esiest to write down everything tht could possibly be subtype nd then cross out the ones tht ren t subtypes. Answer: Recll tht i i f is prsed s i (i f). So, if you work through the rule for function subtyping, you will see tht every combintion of types with the right shpe is possible: i i i i i f i f i i f f f i i f i f f f i f f f b. (7 points) In clss, we rgued it would be unsound to llow int ref flot ref. Demonstrte the issue by writing down, in OCml nottion, progrm tht would type check under this subtyping rule but tht would go wrong with type error t run-time. Explin your nswer very briefly. Answer: let (x:int ref) = ref 0 in let (y:flot ref) = x in y := 3.0; x + 1 (* uh oh! integer rithmetic on flot *) Note: severl people wrote nswers showing how n int could rech n rgument of +. (floting point ddition). However, the premise of this whole question is tht int is subtype of flot, nd it s unresonble to ssume tht would be the cse without lso djusting +. to hndle integers. 4

5 Question 3. Intervl Anlysis (30 points). In this question, we will develop n intervl nlysis, which, for ech vrible x t ech progrm point, determines closed intervl [, b] such tht the run-time vlue of x is gurnteed to be in the intervl [, b]. We lso llow nd b to be nd, respectively, in cse we cnnot bound the intervl on one or both sides. Use for the empty invervl. For exmple, here is CFG nnotted with the intervls determined fter ech sttement (empty intervls omitted): x := 42 x [42, 42] x [45, 45] x := x + 3 x := 5 x [5, 5] y := x - 1 x [5, 45], y [4, 44] x := x - 1 x [-, 44], y [4, ] x [-, 44], y [4, ] y := y + 1 z := x + y x [-, 44], y [4, ], z [-, ] Note tht we hve left out ny conditionl tests; s is usul in dtflow nlysis, your nlysis should lwys ssume ll brnches could be tken.. (5 points) Should the nlysis be forwrd or bckwrd? Answer: Forwrd. b. (5 points) Wht should the initil fcts be t the entry or exit of the progrm? (You cn explin in words.) Answer: Every vrible should be mpped to the empty intervl. c. (5 points) Wht should be in the lttice? (You cn explin in words.) Answer: Every vrible should be mpped to the empty intervl. Mny students nswered tht [, ] should be, but tht doesn t work for two resons. First, mthemticlly, it shouldb be tht x = x. But with the opertion defined below, x would be. Second, dtflow nlysis should strt with the most optimistic ssumption ( ) nd monotoniclly decrese to the ctul nswer. In this cse, it s most optimistic to ssume t progrm point tht ech vrible x mps to the empty intervl, nd then s we discover more informtion bout it, expnd the intervl to contin its ctul vlue. 5

6 d. (5 points) Suppose tht on one incoming edge to join point, x [, b], nd on nother incoming edge to the sme point, x [c, d]. Wht should (x [, b]) (x [c, d]) be defined s? Answer: x [min(, c), mx(b, d)]. e. (5 points) Suppose x [, b] just prior to ech of the following sttements. Write down the new dtflow fct x [c, d] fter the sttements: i. x := 42 Answer: x [42, 42] ii. x := x + 1 Answer: x [ + 1, b + 1] iii. x := 4 - x Answer: x [4 b, 4 ] f. (5 points) If we implement the usul dtflow nlysis lgorithm, is the lgorithm gurnteed to terminte? Why or why not? Answer: No, it s not gurnteed to terminte becuse this lttice does not hve finite height. 6

7 Question 4. Dt flow nlysis (12 points). Here is the control-flow grph from the lst problem gin, this time with numbers for ech sttement: 1. x := x := x x := 5 4. y := x x := x y := y z := x + y. (6 points) Write down the sets of live vribles t the beginning of ech sttement. Write for the empty set, if necessry. Stmt Live vribles t beginning of stmt 1 2 x 3 4 x 5 x,y 6 x,y 7 x,y b. (6 points) Write down the sets of reching definitions t the end of ech sttement. Write for the empty set, if necessry. Stmt Reching definitions t end of stmt ,3,4 5 4,5,6 6 5,6 7 4,5,6,7 7

8 Question 5. Code genertion nd register lloction (24 points). Below is snippet of 08-codegen-2.ml from clss, showing the input expression lnguge, the bytecode instruction lnguge, nd compiltion. We ve mde two smll chnges: We renmed L Register to L Reg to sve some writing; nd we removed reds nd writes through pointers nd identifiers. type expr = EInt of int EAdd of expr expr ESub of expr expr EMul of expr expr EIfZero of expr expr expr type reg = [ L Reg of int ] type src = [ L Int of int ] type instr = ILod of reg src ( dst, src ) IAdd of reg reg reg ( dst, src1, src2 ) IMul of reg reg reg ( dst, src1, src2 ) IIfZero of reg int ( gurd, trget ) IJmp of int ( trget ) IMov of reg reg ( dst, src ) let rec comp expr (st :( string int ) list ) = function EInt n let r = next reg () in (r, [ILod ( L Reg r, L Int n)]) EIfZero (e1, e2, e3) let (r1, p1) = comp expr st e1 in let (r2, p2) = comp expr st e2 in let (r3, p3) = comp expr st e3 in let r = next reg () in (r, [ IIfZero ( L Reg r1, (2+(List. length [IMov ( L Reg r, L Reg r3); IJmp (1+(List.length [IMov ( L Reg r, L Reg r2)] ). (14 points) Suppose we extend the source lnguge with short-circuiting disjunction EOr(e1, e2) tht does the following: First, it evlutes expression e1 to produce vlue v. If v is non-zero, then v is returned s the vlue of the disjunction. Otherwise, expression e2 is evluted nd its vlue is returned. For exmple, EOr (EInt 1, EInt 2) evlutes to 1, nd EOr (EInt 0, EInt 2) evlutes to 2. (Notice tht e2 is not evluted if e1 is non-zero.) Write cse of comp expr tht compiles EOr. let rec comp expr (st :( string int )... EOr (e1, e2) list ) = function Answer: let (r1, p1) = comp expr st e1 in let (r2, p2) = comp expr st e2 in let r = next reg () in (r, [ IIfZero ( L Reg r1, 2); IMov ( L Reg r, L Reg r1); IJmp (1 + (List. length [IMov ( L Reg r, L Reg r2)]) Severl students lso discovered much simpler nswer: comp expr (EIfZero(e1, e2, e1)) (Note tht this is not n idel solution, though, becuse it my evlute e1 twice, which is probbly something we cre bout if we re implementing short-circuiting disjunction.) 8

9 b. (10 points) Finlly, consider the following slight modifiction of the CFG from the erlier problems: x0 := 42 x1 := x0 + 3 x1 := 5 y0 := x1-1 x2 := x1 y1 := y0 x2 := x y1 := y z0 := x2 + y1 Drw the interference grph for the vribles referred to in the bove CFG. After you hve drwn the grph, color it by lbeling nodes with colors, b, c, d, etc, using the miniml number of colors possible. Answer: The nswer we were expecting to get ws the following: x1 x0 b y0 x2 z0 y1 b This nswer ssumes tht in sttement such s x:=y, there is no interference between x nd y which there s not, since the red of y occurs before the write of x. However, severl people ssumed tht x nd y would interfere in tht sttement. Since we weren t completely cler in the lecture notes how to hndle this sitution, we gve lmost full credit for n nswer with tht ssumption, which is the following: b x1 x0 y0 x2 c z0 y1 b 9