《编译原理》考试试题及答案

（(X)（()()()()()()()()()()())和(SS{print"2"}{print"3"}A{print"4"}ABprocedurep(x,y,z)；z:=z*z;P(A,A,B);PrintA,BABSABBA8.写一个文法G,使其语言为L(G)={albmclanbn|l>=0,m>=1,n>=2}ST(),--()--,procedurep(x,y,z);p(a+b,b,a);printaAprocedurep(x,y,z)；p(a+b,a-b,a);printaSFST2.语句ifEthenSSTfori:＝E(1)toE(2)doSi:＝E(1)LIMIT:＝E(2)again:ifi＜＝LIMITtheni:＝i＋1ifc>0thena:=a+1elsea:=a*3-1;STABSTifX>0thenwhileX>0doX:=A*3ETF(1)给出句型(i+i)*i+i的最左推导及画出语法树；STU1.×2.×3.×4.5.6.×7.×8.×9.12.15.17.(初）（终）21.(LL(1)文法）26.(句子)SABCD4.传地址A=6,B=165.L(G)={danbm|n>0,mADBABBAf(),12.正规式a(a|b)*。SB-*/18.(a*b|b*a)={a,b,ab,ba,aab,bbaAC22.逆波兰式abc+e*bc+f/+:=(1)+(2)*(3)+(4)/f(5)+(6):=(1)FIRST((2)如果SFSFSFIRST(S)={a,^,(}，FOLLOW(S)={#,,,)}FIRST(T)={a,^,(}，FOLLOW(T)={}}FIRST(T')={,,SSS2.(1)CSCS3.(1)FIRSTVT(S)={a,(}FIRSTVT(T)={+,aa,(}LASTVT(S)={a,)}LASTVT(T)={+,a,)}(()4.(1)FS(2)F{GEN(:=,E(1).place,_,entry(i));F.place:=entry(i);GEN(:=,E(2).place,_,LIMIT);GEN(jGEN(j,_,_,0)}SGEN(+,F.place,1,F.place);GEN(j,_,_,F.QUAD);S.chain:=F.chain}5.(1)(j<,a,'10',(3))(2)(j,_,_,(12))(3)(j>,c,'0',(5))(4)(j,_,_,(8))(5)(+,a,'1',T1))(6)(:=,T1,_,a)(7)(j,_,_,(1))(8)(*,a,'13',T2))(9)(-,T2,'1',T3)(10)(:=,T3,_,a)(11)(j,_,_,(1))S=(T)=>(T,S)=>(S,S)=>(a,S)=>(a,(T))=>(a,(T,S))=>(a,(S,S))=>(a,(a,S))=>(a,(a,a))((T,S),a)(T,S),a8.(1){backpatch(s1.nextlist,M2.quad);backpatch(E.truelist,M1.quad);S.nextlist=E.falelist;}(2)短语:aAbcde,Ab,d10.(1)S(2)FIRST(S)={a,(}FIRST(S')={a,(,FIRST(L)={a,(}FIRST(L')={,,FOLLOW(S)={,,),#}FOLLOW(S')={,,),#}11.(1)(j>,X,0,(5))(2)(j,_,_,(3))(3)(j<,Y,0,(5))(4)(j,_,_,(11))(5)(j>0,X,0,(7))(6)(j,_,_,(7))(7)(*,A,3,T1)(8)(:=,T1,_,N)(9)(j,_,_,(5))(10)(j,_,_,(13))(11)(+,B,3,T2)(12)(:=,T2,_,Y)=>(F+T)*F+T=>(i+T)*F+T=>(i+F)*F+T=>(i+i)*F+T=>(i+i)*i+T=>(i+i)*i+F=>(i+i)*i+i(2)短语i,F,E+T,(E+T),(E+T)*i,(E+T)*i+F素短语i,E+T13.(1)FIRSTVT(S)={FIRSTVT(T)={FIRSTVT(U)={i,-}LASTVT(S)={LASTVT(T)={LASTVT(U)={i,-}i--B．(xyx)*c．xnyxn(nd．x*yx*12．文法G描述的语言L(G)是指(A)TP17．文法G：STA．{b，C．{b，B．{b，D．{b，TTPB．P,P+T,i,P+T+iB．(a+b)／(c+d)(╳)(╳)((╳)和(U)*也都是正规式，它们所表示的正规集分别为L(U)ABPPbPPbPPbPfPbPS*SS+SS+SS*a<band(c<dore<f)ifa<bgotoL1gotoLfalseL1:ifc<dgotoLtrueL2:ife<fgotoLtruegotoLfalseifa<bthenifc<dthenp:=a+1elsep: