Encoding of Pointers for Hardware Synthesis

Transcription

1 Enong of Ponters for Hrwre Synthess Lu Sémér Govnn De Mhel Computer System Lortory, Stnfor Unversty Stnfor, CA 9405 ABSTRACT In the reent pst, susets of C n C++ hve een efne to moel n synthesze eletron systems s well s reusle IP loks. In orer to synthesze s muh s possle of the C syntx, we hve een reserhng the prolem of syntheszng n optmzng oe wth ponters. In ths pper, we fous on enong the ponters vlues to mnmze the sze of the ruts mplementng the ress trnsltons. After efnng the enong prolem, we present soluton se on heursts n grph-emeng tehnques. The lgorthm hs een ntegrte n our SpC flow whh syntheszes C oe wth ponters. 1. INTRODUCTION 1.1 Synthess from C For yers, esgners hve een wrtng system-level moels usng progrmmng lnguges, suh s C n C++, to estmte the system performne n verfy the funtonl orretness of the esgn. However, to mplement prts of ther esgn n hrwre usng synthess tools, they must mnully trnslte ther oe nto syntheszle suset of hrwre esrpton lnguge (HDL). Ths proess s oth tme onsumng n errorprone. The use of C or suset of C to esre oth hrwre n softwre woul elerte the esgn proess n fltte the softwre/hrwre mgrton. Desgners oul esre ther system n IP omponents usng C. C-se omponent moelng woul mprove esgn reuse n retrgetng to hrwre n softwre mplementtons. However, the tsks of syntheszng hrwre from C turns out to e prtulrly omplte euse of ynm memory lloton, funton ll, reurson, type stng n ponters. In our reserh, we hve een fousng on the prolem of syntheszng oe wth ponters. 1.2 Synthess of Ponters The prolem of syntheszng ponters hs een lrey ntroue n [1] n [6]. The e n [1] s to use ponter nlyss to efne the pont-to set (sets of vrles the ponter my pont to) of eh ponter n the progrm. Then the los (...=*p) n stores (*p=...) n e reple y se sttements n whh the vrles of the pont-to set re expltly referene. Furthermore, to reue the sze of the eoer ssote wth these se sttements n to mnmze the numer of regster use, the vlues of the ponters re enoe. For ponter p, we ll ts enoe vlue ts tg, note p_tg. Exmple 1. Let us onser ponter p tht my pont to the vrles or (result of the ponter nlyss). Conser the followng lne of oe whh mplements lo: out=*p; Ths lo nstruton n e reple y se sttement: swth(p_tg) { se 0: out=; rek; se 1: out=; rek; However, fter enong the ponters vlues n the fnl hrwre mplementton, some omntonl rut my e neee to trnslte the vlues of the ponters nvolve n ssgnments or omprsons. Exmple 2. Conser the ssgnment p=q, where the ponters p n q my pont to or. For p, (resp. ) s ssote wth the vlue 0 (resp. 1), wheres for q, (resp. ) s ssote wth 1 (resp. 0). After enong, the ssgnment p=q s reple y the followng oe segment: swth(q_tg) { se 0: p_tg = 1; rek; se 1: p_tg = 0; rek; In ths se, etter enong n e foun y hvng the sme enong for oth p n q. Then p=q woul retly e reple y p_tg=q_tg. To mnmze the orresponng omntonl rut, the oes hve to e equl or sufel of one nother. Ths pper presents soluton for the enong of ponters vlues. In Seton 2, we re gong to formulte the prolem. In orer to reue the omplexty, we show n Seton how the enong prolem n e smplfe n solve usng grph-emeng tehnques. In Setons 4 n 5, we present two optmzton tehnques lle folng n splttng. The lgorthm s then esre n Seton 6. Fnlly, n Seton 7, we show how the enong of ponters hs een ntegrte n our tool for the synthess of ponters n C (SpC) n present some results. 2. ENCODING OF POINTER 2.1 Defnton of the Prolem Defnton 1. We efne ponter-epenene grph s grph n whh the noes re the ponters n the eges re the reltons etween the ponters. An ege etween two noes s efne when the two orresponng ponters re ssgne or ompre.

2 Exmple. Conser the followng oe segment: nt *, *, *, *q1, *q2;... f(==0) { =&; =&; =&; else { =&; =&; =&; f(j==0) { q1=; q2=; else { q1=; q2=;... In ths exmple, we onser the ponters {,,, q1, q2 n the vrles {,,,. The ponters re efne s follows: my pont to the vrles or, my pont to the or n my pont to or. Then, q1 my tke the vlue of or n q2 my tke the vlue of or. Consequently, q1 ponts to, or, n q2 ponts to, or. Ths les to the ponter-epenene grph on Fgure 1. {, {, {, q1 {,, q1 q2 q2 {,, Fgure 1: exmple of ponter-epenene grph Our gol s to enoe eh ponter wth the mnmum numer of ts. An, when ponter s ssgne or ompre to nother ponter, we wnt the orresponng tgs to e equl (e.g. p_tg=q_tg) or s lose s possle to eh other. If two tgs hve fferent numer of ts, one tg n e equl to sufel of the other. Assgnments woul then e performe y ontentng or removng ts, wheres omprsons woul only e exeute on the ts ommon to the two oes. Ths reues the sze of the rut tht trnsltes or ompres the tgs whle keepng the numer of ts to mnmum. The enong prolem n e formulte s follows. For eh ponter we efne set of symols orresponng to the vrles the ponter my pont to. As result we hve n ensemle of sets of symols n the epenenes mong the sets. The prolem onssts of enong the symols n the sets. The onstrnts on the enong re two: 1) the superue 1 of the symols n eh set must hve mnmum sze; 2) the symols tht orrespon to the sme vrle n two epenent sets must e enoe s lose s possle. The reson for the frst onstrnt s to mnmze the numer of ts to store, whle the reson for the seon one s to reue the sze of the omntonl rut mplementng the ponter ssgnment n omprson. Exmple 4. In Exmple, the ponters, n my pont to two fferent vrles n q1, q2 my pont to three fferent vrles. We wnt to oe, n on 1 t n q1, q2 on 2 ts. Then we wnt the oe of n to e sufels of the oe of q1 n the oe of n to e sufels of the oe of q2. An enong verfyng these propertes s shown on Fgure Fgure 2: exmple of enong 1. The superue of set of ues s the smllest ue ontnng ll the ues n the set [8]. q1,q2 For, vlue 0 s ssgne to n vlue 1 to. For, 1 wll e ssgne to n 0 to. As result, q1= wll e reple y q1_tg={_tg,0 n q1= wll e reple y q1_tg={0,_tg (where {, s the ontenton opertor). 2.2 Prolem Formulton Let s efne P ponters { p 1, p 2,, p P. We efne the ponter epenene relton s follows. For two ponters, p n p j, r( p, p j ) = 1 f n only f the two ponters re ssgne or ompre, r( p, p j ) = 0 otherwse. For eh ponter p we efne ts pont-to set Π to e the set of vrles the ponter my pont to. The pont-to set Π s set of N symols { s 1, s 2,, s N, where eh symol s ssote wth vrle. After enong, we efne E, the set of the enoe symols of the pont-to set Π. The enoe vlues of the symols n eh set re note { e 1, e 2,, e N. Defnton 2. Two sets Π n Π j re s to e epenent f ther ssote ponters re ssgne or ompre (.e. r( p, p j ) = 1 ). Our frst gol s to mnmze the numer of regsters s well s the sze of the eoers requre to store n eoe the ponters vlues. We wnt the mnmze the menson of the superue of the enoe symols n eh set. Ths mnmum s heve when the sum of the mensons of the superues s lso mnmze: P mn m( superue( E )) (1) = 1 When two ponters re ssgne or ompre, we lso wnt to mnmze the sze of the rut mplementng the trnslton of the oes. For ths purpose, the stne etween enoe symols n two epenent sets hs to e mnmum: P P mn r( p, p j )st( E, E j ) (2) = 1 j = 1 where st( ) s the stne etween the two enoe sets. When the ponters hve the sme pont-to set, st( ) s efne s: N j st( E, E j ) = H( e k, e k ) () k = 1 where N s the numer of symols n the pont-to set n j H( e k, e k ) s the Hmmng stne etween the oes of the j symols s k = s k. In generl, the szes of the oes n the pont-to sets my ffer. The omputton of the stne s then more omplex: we hve to onser the fferent sufels n the symols ssote wth the sme vrles n the two sets. Our gol s to mnmze Eq. 1 n Eq. 2. There s tre-off etween the storge re (numer of regsters) n the mount of log use to trnslte the oes. For exmple, one my optmze the sze of the ponters keepng the mount of log mnmum y mnmzng frst Eq. 2 n then Eq. 1. In generl, we n st the prolem s follows: P P P mn β m( sup. ( E )) + γ r( p, p j )st( E, E j ) (4) = 1 = 1 j = 1 where β n γ re two oeffents.

3 Sne ths prolem s omputtonlly hr to solve usng ext methos, we hve een lookng t heursts for solvng the enong of ponters.. SIMPLIFIED PROBLEM.1 Formlsm for Glol Soluton In the generl formulton of the prolem presente n Seton 2, fferent oes my e ssote wth the symols n eh set. Therefore the enong hs to e foun lolly, n eh set. The prolem n e smplfe y onstrnng ll symols ssote wth the sme vrle to shre the sme oe. Eq. 2 s then rrelevnt euse the stne etween the oes of the symols tht orrespon to the sme vrle n the fferent pont-to sets s null (.e. st( E, E j ) = 0 (, j) { 1, 2,..., P 2 ). Our gol eomes to mnmze Eq. 1 only. The enong s then foun glolly for ll the symols whh orrespon to the sme vrle n the sets. We re now onserng the symols (.e. vrles) n the unon Π of the pont-to sets. These symols wll e enote: Π = { s 1, s 2,, s N. The omplexty s reue: nste of elng wth O(P*N) symols { s j suh tht { 1, 2,..., P n j { 1, 2,..., N we only el wth N symols { s 1, s 2,, s N, where N s the numer of vrles. We ntroue now formlsm whh hs een use to solve other enong prolems [8], [7], [14]. Defnton. The relton mtrx A s efne s the mtrx n whh the rows represent the pont-to sets n the olumns the symols. The entry, j of A s 1 f n only f the symol s j s n the set Π. Exmple 5. Let s tke the se of Exmple. We n onstrut the followng relton mtrx: q1 q2 For exmple, the frst row of the mtrx shows tht my pont to or. We re lookng for n enong mtrx E, tht stsfes the enong onstrnts represente y A. The onstrnts expresse y the relton mtrx A s the followng. Eh row n A orrespons to pont-to set. For eh row α of A, we wnt the superues of the rows of E orresponng to the 1s n α to hve mnmum sze. Ths orrespons to the onstrnt expresse n Eq. 1. Ths prolem ffers from the nput enong prolem [8]. The relton mtrx s not nry onstrnt mtrx, s efne for the generl enong prolem [8], n the sense tht we on t hve the tonl onstrnt tht, for eh row α of the onstrnt mtrx, the superue of the rows of E orresponng to the 1s n α oes not nterset ny of the rows of E orresponng to the 0s n α. The 0s n the relton mtrx n then e onsere s on t re n the onstrnt mtrx..2 Generl Enong Algorthm The prolem of nput enong hs een extensvely stue ([1] [4] [7] [9] [10] [11] [12] [14]). We use n pproh remnsent of MUSTANG [9] n POW [1]. Defnton 4. An ffnty grph s n unrete weghte grph n whh the noes re the symols Π = { s 1, s 2,, s N n the eges re the reltons etween the symols n Π. The weght w, j on the ege { s, s j s efne s: P w, j = k, ( 1 + k, j Log 2 N Log 2 N k ) (5) k = 1 where P s the numer of ponters, N s the totl numer of symols, N k the numer of symols n the set Π k, n, j s n element of the relton mtrx. The weght w, j n the ffnty grph nreses wth the numer of sets tht ontn oth s n s j : when two vrles re n mny pont-to sets, we wnt ther oe to e lose. Ths s even more mportnt for smll pont-to sets. For exmple, f we hve N k = 2 symols n the pont-to set Π k, ther oe must e next to eh other to mnmze the menson of the superue of the enoe set E k, wheres f we hve N k = 10 symols n the pont-to set Π k, the Hmmng stne etween the enong of the symols n the pont-to set n e s muh s Log 2 ( N k ) = 4. Therefore, the weght w, j s the sum of the ontrutons of the pont-to sets tht ontn oth s n s j, where the ontruton of eh pont-to set Π k s ( 1 + Log 2 N Log 2 N k ). The ponter enong prolem n e solve s n emeng of the ffnty grph n the Boolen hyperspe s one n [10],[9],[1]. Exmple 6. The relton mtrx presente n Exmple 5 n e use to generte the ffnty grph on Fgure Fgure : Exmple of Affnty Grph Let s look t w, the weght on the ege {,. The vrles n re oth n the pont-to sets of n q1. The weght w, s, sum of 2, ontruton from, n 1, ontruton from q1. After grph-emeng, the enong presente n Fgure 2 n e foun. These lgorthms however o not onser the ft tht two symols n shre the sme oe. We re gong to use ths property n Seton 4 for tehnque lle folng. One symol n lso hve multple oes. The noton of splttng presente n Seton 5 wll e se on ths property. 4. ENCODING WITH FOLDING Defnton 5. We efne s folng the ton of ssgnng the sme oe to two symols. Proposton 1. Two symols n e fole f n only f they re not oth n the sme pont-to set n not n two epenent pontto sets. The rtonle for ths proposton s tht we wnt to stngush eh symol nse pont-to set n, n the se of omprson, we wnt to stngush the symols n the two epenent pont-to sets. In the relton mtrx A, folng the symols s n s j s equvlent to replng the olumns n j y one olumn k suh tht: k, l =, l jl, for l n {1, 2,..., N. (6)

4 In the ffnty grph, folng s one y mergng (or fusng 1 ) the noes orresponng to the symols s, s j nto one new noe orresponng to s k. The weghts on the eges nent to ths new noe orresponng to s k re then efne s: w k, l = w, l + w jl, for l n {1, 2,..., N. (7) Grph-emeng tehnques n e mofe to norporte folng. In Seton 6, we present olumn-se metho wth folng. Exmple 7. Let s onser the ponter-epenene grph on Fgure 4 where,, n pont respetvely to {,,, {,, n {,,e. The relton mtrx n the ssote ffnty grph re represente n Fgure 5. {,, {,, {,,e q1 {,,, q1 q2 q2 {,,,e Fgure 4: Ponter-epenene grph e q q2 Fgure 5: Enong prolem efore folng The numer of vrles n eh pont-to set s ether (for,, n ) or 4 (for q1 n q2). Therefore, we wnt to oe the symols ssote wth the vrles on 2 ts. However, sne we hve 5 symols, n enong wth less thn ts nnot e foun wthout folng. The symol s n the pont-to set of n q1, wheres the symol e s the pont-to set of n q2. Aorng to the ponter-epenene grph, these pont-to sets re not epenent. The symols ssote wth n e n e fole. After folng we en up wth the grph on Fgure 6. e q1 q2,e Fgure 6: Enong prolem fter folng n e Ths les to n enong tht requres only two ts: 00,, 01 q1,q e 00,e Fgure 7: Result of the enong fter folng n e 1. A pr of vertes, n grph re s to e fuse (merge or entfe) f the two vertes re reple y sngle vertex suh tht every ege tht ws nent on ether or or on oth s nent on the new vertex [2] e 5. SPLITTING TECHNIQUE Defnton 6. We efne s splttng the ton of ssgnng two or more oes to one vrle (or symol). In Seton n 4, eh vrle ws ssote wth unque symol tht ws enoe. After splttng, one vrle my e ssote wth more thn one symol: splttng symol s s equvlent to retng new symol s ' whh orrespons to the sme vrle. The orgnl symol s n the newly rete s ' re then enoe nto e n e ' respetvely. Proposton 2. A pont-to set Π k tht ontns symol s my, fter splttng s, ontn the newly rete symol s ' f n only f there s no oe equl to e ' n the enoe set E k or n ny enoe set epene of Π k. As esre n Seton 2.2 the symols n eh set n hve fferent oes. Therefore, to mnmze the menson of the superue of the enoe symols n set (.e. Eq. 1), we n rete new symols ssote wth the sme vrles, for ths set. Note tht, f we splt the symols for eh pont-to set, we en up wth lol enong sheme s presente n Seton 2.2. However, to lmt the nrese n omplexty, we re tryng to splt s few symols s possle. In the relton mtrx A, splttng s one y ng olumn ' reltve to s '. For eh row α reltve to the pont-to set ' k Π k, the entry k n e set to 1 f Proposton 2 s verfe. The pont-to set Π k my then ontn s, s ' or oth s n s '. In orer to mnmze Eq. 1, for set tht my ontn s or s ', the followng expresson my e onsere: mn( m( superue( ( E k { e ) E' ))) (8) E' where E' s ether { e, { e ' or { e, e '. For exmple, f Eq. 8 s mnmum for E' = { e ', the menson of the superue of the enoe symols n the pontto set Π k s mnmum when k s set to 0 n k s set to 1. ' The new ffnty grph n then e reompute from the relton mtrx. Splttng s well s folng n e norporte n our grph-emeng lgorthm s presente n Seton 6. Exmple 8. Let s onser the ponter-epenene grph on Fgure 8 where, n my respetvely pont to {,, {, n {,. {, {, q {, q {,, A = q Fgure 8: enong prolem efore splttng We woul lke to enoe, n wth 1 t n q wth 2 ts. We lso wnt the oes of, n to e sufels of the oe of q. Usng the enong tehnque wthout splttng symols, we n fn the enong on Fgure 9. In ths se n re enoe on 1 t ut the enong of requres 2 ts.

5 00 q, Fgure 9: enong wthout splttng After splttng the symol, we en up wth the two symols n. We n fn the enong on Fgure 10 where the symol s n the pont-to set of, n q n n the pont-to set of n q. A = ' The enong n Fgure 10 s optml:, n re enoe on 1 t n the ssgnments to q (q=, q=, q=) on t requre ny tonl log. 6. ALGORITHM q q 11 Fgure 10: result fter splttng symol We propose olumn-se pproh (whh mens tht the enong mtrx n e foun olumn y olumn []). Our lgorthm wthout folng n splttng s smlr to the one use n [1]. A pseuo oe of the lgorthm wth folng n splttng s presente on Fgure 11. We onser one t of the oe t the tme. For symol s k ssote wth the oe e, we onser the ts e for k={1, 2,..., n where n s typlly n = log. At eh terton k, we onstrut the k th 2 ( N ) olumn of the enong mtrx E: the vlue k of e (0 or 1) s ssgne n wy tht mnmzes the stne etween the enoe symol e n ts neghors n the ffnty grph. The ssgnment s one for the symols on every ege strtng wth the eges wth hghest weghts. For the symols nent to the eges { s, s, we re tryng to ssgn the sme vlue j to oth e n e j. However, f the symols s n s jw re lso nent to other eges whose weghts re hgher thn, the, j vlues of e n e j my lrey e ssgne to fferent vlues. Moreover, t eh terton of k, the numer of symols hvng the sme oe s lso lmte. Defnton 7. An ege { s, s s s to e volte t terton k j f the vlues e n e j ssote wth the two symols nent to the ege, hve fferent vlues. Defnton 8. A lss volton s efne t terton k, when more thn 2 n k symols hve the sme oe so fr. (At terton k, we re only onserng the k frst ts of the oes, sne the other ones hven t een ssgne yet). The rtonle for efnng lss volton s tht we ultmtely wnt to stngush ll the symols. Therefore, n our greey lgorthm, we hve to mke sure tht, t eh terton of k, we hve less thn 2 n k symols ssote wth the sme oe. For exmple for k=(n-1), we nnot hve more thn 2 symols wth the sme oe. Proposton. An ege { s, s j s volte t terton k f ether one of the followng ontons pples: there s lss volton (n therefore, e n e j nee to hve fferent vlues), the vlues e n e j ssote wth the two symols nent to the ege hve lrey een ssgne to fferent vlues (y Defnton 7). In the se of lss volton, we try to fol one of the symols on the ege { s, s j wth ny of the prevously ssgne symols. At ths stge, two symols re fole f Proposton 1 s verfe n f they hve the sme prtl oe so fr. In the se when fferent vlues hve een ssgne to the two symols s, s j (.e. e e j ), we try to splt the symols on the ege. One symol n e splt f the newly rete symol oes use ny lss volton or n e fole wth nother symol. In our lgorthm, for symol s, we rete new symol l l s ' ssote wth oe e ' suh tht e' = e for l>k n e' = e j. In se of lss volton, we try to fol ths new symol s one prevously. If folng nnot e one, the symol s not splt. enoe ponter() { /* onstrut mtrx E one olumn t tme */ for k=1 to n ssgn_oe(k); ssgn_oe(k) { sort eges y weght n eresng orer; foreh ege { s, s j { f( e n e j not ssgne) { e = e j = selet_t( s, s j ); f(lss volton) { try_fol( s ); try_fol( s j ); f( s or s j not ssgne) { s h =unssgne( s, s j ); s l =ssgne( s, s j ); e h = e l ; f(no lss volton) { try_fol( s ); try_fol( s j ); else /* s n s ssgne */ j f( e h!= e l ) volte_eges->({ s, s j ) sort volte eges y weght n eresng orer; foreh volte ege { s, s j { try_splt( s ); try_splt( s j ); try_splt( s ) { rete s ' e ' = e xor(1<<k); f(lss volton) try_fol( s ' ); try_fol( s ) { f( s j s.t. Proposton 1 verfe n e == e j ){ fol( s, s j ); remove s ; Fgure 11: lgorthm wth splttng n folng

6 Exmple 9. Conser the prolem presente on Fgure 12. The ssote relton mtrx n ffnty grph re presente on Fgure 1. r4 r5 {, {, q1 q2 {, r4 {, r5 {, Fgure 12: enong prolem r4 r5 q1 q2 Fgure 1: relton mtrx n ffnty grph Sne we hve 4 symols, we wnt to enoe them on n=2 ts. At terton k=1, we ssgn the vlue 0 to, n 1 to,. The volte eges re {,, {,, {, n {, ut folng n splttng nnot e one. For exmple, for the ege {,, nnot e fole wth (resp. ) euse oth symols re n the pont-to set of n q1(resp. n q1). At terton k=2, we ssgn 00 to, 01 to, 10 to n 11 to. All the eges re volte n some symols n e splt. The vrle n e splt n the new symol n e fole wth. The vrle n lso e splt n the new symol n e fole wth. Fnlly, we en up wth the enong on Fgure 14 n whh ll the onstrnts re verfe. q1,q2 7. IMPLEMENTATION IN SPC In [1], we presente SpC soluton for the synthess of ponters n C. The toolflow s presente on Fgure 15. Our mplementton tkes funton wth ponters n C n genertes moule n Verlog. Ths moule n then e synthesze usng the Behvorl Compler TM of Synopsys. The trnslton from C to Verlog onssts of fferent psses. After the front-en, we nlne the funtons n perform the ponter nlyss. Then the lsng nformton s use to remove n optmze ponters n the followng orer: - efne the pont-to-set of eh ponter; - reple the los n stores; - optmze los n stores; - enoe ponters; - e-oe elmnton. The ntermete oe wthout ponters s then trnslte nto Verlog ,, 01,r4,r5 1- Fgure 14: enong fter splttng n folng C funton Verlog Moule Netlst Toy, SpC supports only ponters to vrles n rry elements. Nevertheless, our lgorthm for ponter enong oul e use for ponters to ponters, ponters to funtons, n reursve t strutures s well. The heurst lgorthm esre n Seton 6 hs een mplemente n pple to some test ses. The results re llustrte n Tle 1 n hve een otne s follows. The regster fle s well s the log neessry to trnslte the vlues of the ponters hs een synthesze usng Synopsys Desgn Compler TM n the tsm.5 lrry. We present the results for three fferent shemes. Frst we present the results for strghtforwr mnmumlength enong of the symols. In ths suoptml enong, eh vrle n eh pont-to set s smply ssote wth numer (0 for the frst vrle, 1 for the seon vrle, et...). The numer of ts use to enoe eh tg s then mnmum ut the sze of the rut whh trnsltes the vlues of the ponters s not. The seon sheme s the mplementton of the lgorthm wthout splttng n folng. The sze of the rut trnsltng the vlues of the ponters s then reue. However, the numer of ts use for the enong s not lwys mnml, whh les to lrger eong ruts (omntonl re) n more regsters (non-omntonl re). Fnlly, the lst olumn shows the results for the lgorthm wth folng n splttng. The length of the oes s then lose from the mnmum n the sze of the omntonl rut s reue, whh gves etter results. exmple P N. front-en ponter nlyss remove ponters eoe Csuf2Verlog.v Behvorl Compler. set pont-to-set reple * opt lo/store enoe ponters remove & Fgure 15: Toolflow for the Synthess of Ponters n C mn. length smple lgorthm spt n fol omn. non-. omn. non-. omn. non-. test test test Tle 1: results fter synthess n optmzton usng tsm.5 lrry: omntonl re n non-omntonl re n lrry unts.

7 8. CONCLUSION We hve presente the prolem of enong ponters for the synthess of hrwre. After smplfton, the prolem n e solve usng grph-emeng tehnques. These tehnques n e further optmze: two symols n shre the sme oe (folng) n one symol n hve multple oes (splttng). The lgorthm presente hs een suessfully mplemente for the synthess of C oe wth ponters. Our tehnque oul lso e pple to other lnguges n ojet-orente progrms. 9. ACKNOWLEDGMENTS Ths reserh s supporte y DARPA uner grnt DATB6-95-C0049 n y Synopsys In. We woul lso lke to thnk Murzo Dmn for ts preous omments n suggestons. 10.REFERENCES [1] L.Benn n G.De Mhel, Stte ssgnment for low power sspton Custom Integrte Cruts Conferene, 1994, pp [2] Nrsngh Deo Grph Theory wth ppltons to Engneerng n Computer Sene, Prente-Hll, Englewoo Clffs, NJ, [] T.A Dolott n E.J. MCluskey. The Enong of Internl Sttes of Sequentl Mhnes. n IEEE Trnstons on Eletron Computers, volume EC-1, pp , Otoer [4] C. Duff, Coge utomtes et théore es ues ntersetnts, These, Insttut Ntonl Polytehnque e Grenole, Frne, Mrh [5] Evgue Golerg, Tzno Vll, Roert Bryton, Alerto Sngovnn-Vnentell, Theory n Algorthms for Fe Hyperue Emeng Trnston on CAD, volume 17(6), June [6] Hwyong Km, Kyoung Cho, Trnsformton from C to Syntheszle VHDL n Proeengs of As Pf Conf. on HDL APCHDL 98, July [7] Govnn De Mhel. Symol Desgn of Comntonl n Sequentl Log Cruts Implemente y Two-Level Log Mros. IEEE Trnston on CAD, volume 5(4), pp , [8] Govnn De Mhel Synthess n Optmzton of Dgtl Cruts, M Grw Hll, Hghtstown, NJ, [9] A. R. Newton, S. Devrs, H-keung M n A. Sngovnn- Vnentell. MUSTANG: Stte Assgnment of Fnte Stte Mhnes Trgetng Multlevel Log Implementtons. IEEE Trnston on CAD, volume 7(12), pp , [10] G. Suer, Stte Assgnment of Asynhronous Sequentl Mhnes usng Grph Tehnques, IEEE Trnston on Computer, Mrh [11] G. Suer, C. Duff, F. Porot, Stte ssgnment usng new emeng metho se on ntersetng ue theory, n Proeeng of Desgn Automton Conf., pp , June [12] G. Suer, M.C. Depulet n P. Sr, ASYL: A rulese system for ontroller synthess, IEEE Trnstons on CAD, volume CAD-6, pp , Novemer [1] Lu Sémér, Govnn De Mhel, SpC: Synthess of Ponters n C. Applton of Ponter Anlyss to the Behvorl Synthess from C, proeeng of the 1998 ICCAD, pp , Novemer [14] Tzno Vll, Alerto Sngovnn-Vnentelly, NOVA: Stte Assgnment of Fnte Stte Mhnes for Optml Two- Level Log Implementton, IEEE Trnstons on Computer-Ae Desgn, Vol. 9, pp , Septemer 1990.