Compilers. Chapter 4: Syntactic Analyser. 3 er course Spring Term. Precedence grammars. Precedence grammars

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1 Complers Chpter 4: yntt Anlyser er ourse prng erm Prt 4g: mple Preedene Grmmrs Alfonso Orteg: nrque Alfonse: Introduton A preedene grmmr ses the nlyss n the preedene reltonshps mong the symols n the lnguge. Usng those reltonshps, the prser wll try to dede whh s the symol wth the hghest preedene, to strt uldng the ottom-up prse. Prevous exmple he purpose s to uld the grmmr n suh wy tht some symols hve preedene over other symols. or nstne, the followng grmmr does not gve preedene of over : ::= ut the rule ::= hs preedene over the other two rules, euse they nnot pply f the s hven t een redued to s 4

2 Prevous exmple Prevous exmple 5 6 Prevous exmple Ide In the followng sldes we shll formlse some reltons of preedene etween the symols of the grmmr: Whh symols hve equl preedene. Whh prs of symols re suh tht the frst one hs more preedene thn the seond one. 7 8

3 HA reltonshp efnton HA reltonshp trx representton Σ Σ N Let G={Σ N,Σ,P,Α} e ontext-ndependent grmmr. If U s non-termnl symol n the grmmr, we shll ll hed(u) the set of symols tht n e t the egnnng of the strngs derved from U. If there s rule U ::= Vx x...x n, V hed(u) If V hed(u), nd, W hed(v), then W hed(u) ::= ::= ::= ::= ::= ::= () ( ) ( ) 9 AIL reltonshp efnton AIL reltonshp trx representton Σ Σ N Let G={Σ N,Σ,P,Α} e ontext-ndependent grmmr. If U s non-termnl symol n the grmmr, we shll ll tl(u) the set of symols tht n e t the end of the strngs derved from U. If there s rule U ::= x x...x n V, V tl(u) If V tl(u), nd, W tl(v), then W tl(u) ::= ::= ::= ::= ::= ::= () ( ) ( )

4 Preedene reltonshps wo symols R nd, n grmmr, hve equl preedene () f nd only f there s ny rule n the grmmr n whh they pper onseutve R f nd only f there s rule (U ::= xry) Preedene reltonshps A symol R hs less preedene thn symol (R ) f nd only f there s rule n the grmmr U ::= w Rw, where w,w (Σ N Σ ) nd Hed() U U... R R... Z 4 Preedene reltonshps A symol R hs more preedene thn symol (R ) f nd only f there s rule n the grmmr U ::= w Ww, where w,w (Σ N Σ ) R l() ther Hed(W), or =W. U U Preedene reltonshps A grmmr s smple preedene grmmr f, for every pr of symols, there s only one preedene reltonshp. If we hve sententl form, we n lulte the preedene reltonshps etween eh pr of onseutve symols: x x x x4 x5 x6 x7 x8 x W... A hndle wll lwys strt wth reltonshp, wll possly ontnue wth ny numer of reltonshps, nd wll lwys end wth reltonshp. We re gong to see now how to uld the preedene mtrx showng the reltonshps.... R... R

5 xmple: reltonshp xmple: HA reltonshp ::= ::= ::= ::= 7 8 xmple: AIL reltonshp xmple: reltonshp ::= ::= x = It n e lulted s the (Boolen) mtrl produt Hed ::= ::= 9

6 xmple: dentty reltonshp xmple: reltonshp ::= ::= x x = It n e lulted s the (Boolen) mtrl produt: (AIL) t (HA OR Identty) ::= ::= Preedene reltonshps Preedene reltonshps ::= ::= ::= ::= If we hve the word: the reltonshps re: wth the hndle mrked n red. 4

7 Preedene reltonshps Prsng lgorthm N ::= ::= AR he mtrx n e extended wth AR nd n N symol Prse. Intlse stk wth the AR symol.. Compre top(stk) wth the next element n the strng, X.. If top(stk) s not relted to X, notfy syntt error.. If top(stk) X, push(x). If top(stk) X, push(x) 4. If top(stk) X, pop from the stk untl we fnd the reltonshp. he symols extrted from the stk re the hndle. We look for the hndle n the rght-hnd-sdes of the grmmr rules, nd susttute t y the left-hnd-sde, whh wll e pushed n the stk. he reltonshps wll e: AR N wth the hndle mrked n red. 5 6 Prsng lgorthm Prsng lgorthm N N N N A R,< A R AR AR 7 8

8 9 N AR N Prsng lgorthm A R,<,< N AR N Prsng lgorthm A R,<,<,< N AR N Prsng lgorthm A R,<,< N AR N Prsng lgorthm A R,<,<,=

9 N AR N Prsng lgorthm,=,<,< A R, = 4 N AR N Prsng lgorthm A R,< 5 N AR N Prsng lgorthm A R,<,= 6 N AR N Prsng lgorthm A R,<,=,=

10 7 N AR N Prsng lgorthm A R 8 N AR N Prsng lgorthm A R,< ACCP 9 N AR N Prsng lgorthm A R 4 N AR N Prsng lgorthm A R,<

11 Prsng lgorthm mplfton wth preedene funtons N N AR,<,< A R RROR: ther s no preedene etween nd, so the word s rejeted. 4 he preedene mtrx hs sze of N y N. We n represent t wth less spe usng ouple of funtons, f nd g, lled preedene funtons. hey hve the followng propertes: f(x) = g(y) f nd only f x y f(x) < g(y) f nd only f x y f(x) > g(y) f nd only f x y xmple: N f g AR - N - AR 4 mplfton wth preedene funtons Proedure for lultng the preedene funtons Advntges: he reltons of preedene n e stored n mtrx of sze xn, rther thn NxN. sdvntges: We lose some nformton, suh s the empty ells n the orgnl mtrx. In some ses, t wll e mpossle to fnd suh funtons. or nstne, onsder the followng mtrx: he restrtons re: f() = g() f() > g() f() = g() f() = g(), whh re mpossle to stsfy. he preedene funtons n e lulted wth the followng proedure:. Represent n dgrm: unton f, ppled to ll the symols n the grmmr, n row. unton g, ppled to every symol, n row elow.. or every pr (x,y), rw n r from f(x) to g(y) f x y or x y rw n r from g(y) to f(x) y x y or x y. nlly, for every ell f(x) or g(x): We ount how mny ells re essle from t, n zero, one or severl steps. ht numer wll e the vlue of the funton for tht symol. 4 44

12 Proedure for lultng the preedene funtons Proedure for lultng the preedene funtons f() f() f() f() f() f() f() f() g() g() g() g() g() AR g() g() g() N 45. We strt from f(). here re two nodes esle (nludng tself): f() nd g(). herefore, we set f() =. rom f(), we n go to f() nd g(). herefore, we set, gn, f() =. rom f(), we n go to {f(), g(), f()} -> f() = 4. rom f(), we n go to {f(), g(), f()} -> f() = 5. rom g(), we n go to {g(), f()} -> g() = 6. rom g(), we n go to {g(), f(), g()} -> g() = 7. rom g(), we n go to {g(), f()} -> g() = 8. rom g(), we n go to {g(), f(), g()} -> g() = 46 Proedure for lultng the preedene funtons Proedure for lultng the preedene funtons If x y, then f(x) = g(y), euse every node essle from f(x) wll e essle from g(y). If x y, then here s n rrow from f(x) to g(y), ut there s not n nverse rrow. herefore, we nnot go k from g(y) to f(x). hs mens tht from f(x) we ll e le to ess every node essle from g(y), plus f(x), plus every other node essle from f(x). f(x) wll e strtly hgher thn g(y) nlly, f x y, then, here s n rrow from g(y) to f(x), ut not the nverse. herefore, the numer of nodes essle from g(y) wll e strtly hgher thn the numer essle from f(x) -> g(y) > f(x). We n utomte the proedure n the followng wy:. Buld the mtrx B: (<=) t >=. Clulte the mtrx B = B OR B OR B OR B OR.... f(x) wll e the ddton of eh row n the top hlf of the mtrx. 4. g(x) wll e the ddton of eh row n the lower hlf of the mtrx

13 49 Proedure for lultng the preedene funtons N trt >= <= 5 Proedure for lultng the preedene funtons >= <= B 5 Proedure for lultng the preedene funtons B 5 Proedure for lultng the preedene funtons B g() g() g() g() f() f() f() f() Σ Σ Σ Σ Σ Σ Σ Σ

14 Complete exmple xmple: HA reltonshp = Hed ::= ::= ::= ::= ::= ::= 5 54 xmple: AIL reltonshp xmple: reltonshp ::= ::= ::= l = H x = < Hed 55 56

15 xmple: dentty reltonshp xmple: reltonshp ::= ::= ::= Id t = H x x = > (AIL) t (HA OR Identty) Preedene reltonshps Prsng lgorthm ::= ::= N A R N t 59 6

16 6 N Prsng lgorthm A R,< t N 6 N Prsng lgorthm A R t N 6 N Prsng lgorthm A R,< t N 64 N Prsng lgorthm A R,<,= t N

17 65 N Prsng lgorthm A R,<,=,< t N 66 N Prsng lgorthm A R,<,= t N 67 N Prsng lgorthm A R,<,=,< t N 68 N Prsng lgorthm A R,<,= t N

18 69 N Prsng lgorthm A R,<,=,= t N 7 N I Prsng lgorthm A R t N 7 N Prsng lgorthm A R,< t N 7 N I Prsng lgorthm A R t N

19 7 N Prsng lgorthm A R,< t N 74 N Prsng lgorthm A R,<,= t N 75 N Prsng lgorthm A R,<,=,< t N 76 N Prsng lgorthm A R,<,= t N

20 77 N Prsng lgorthm A R,<,=,< t N 78 N Prsng lgorthm A R,<,= t N 79 N Prsng lgorthm A R,<,=,= t N 8 N Prsng lgorthm A R t N

21 Prsng lgorthm Complete exmple N N Note tht the followng versons of the sme grmmr re not smplepreedene grmmrs: A R ::= ::= ::= ACCP ::= ::= ::= t Why? (exerse) 8 8 Proedure for lultng the preedene funtons Complete exmple f() f() f() f() f() f() unton f() Aessle ells f(), g() unton vlue f() f(), g(), f() f() f(), g(), f(), g() 4 g() g() g() g() g() g() f() f() f(), g(), f() f(), g(), f() 8 f() g() g() g() g() g() g() f(), g(), g(), f(), f() g() g(), f(), f() g(), f(), f(), g() g(), f() g(), f(), g(), f() g(), f(), f(), g()

22 Prsng lgorthm unton f() unton vlue N f() f() f() 4 A R f() f() g() g() g() g() 5 4 XRCI: Repet the nlyss of the strng usng the preedene funtons g() 4 g() 4 f(ar)=g(n) 85