编译原理第三版课后习题答案.doc

该【编译原理第三版课后习题答案 】是由【花双韵芝】上传分享，文档一共【28】页，该文档可以免费在线阅读，需要了解更多关于【编译原理第三版课后习题答案 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。精品文档目录P36-6 2P36-7 2P36-8 2P36-9 3P36-10 3P36-11 3P64–7 4P64–8 5P64–12 5P64–14 7P81–1 8P81–2 9P81–3 12P133–1 12P133–2 12P133–3 14P134–5 15P164–5 19P164–7 19P217–1 19P217–3 20P218–4 20P218–5 21P218–6 22P218–7 22P219–12 22P270–9 -6(1)L(G1)是0~9组成的数字串(2)最左推导:NNDNDDNDDDDDDD0DDD01DD012D0127NNDDD3D34NNDNDDDDD5DD56D568最右推导:NNDN7ND7N27ND27N127D1270127NNDN4D434NNDN8ND8N68D68568P36-7G(S)1|3|5|7|9N2|4|6|8|OD0|NSO|AOAAD|NP36-8文法:E T|E T|E TF|T*F|T/FF(E)|i最左推导:EETTTFTiTiT*FiF*Fii*Fii*iETT*FF*Fi*Fi*(E)i*(ET)i*(TT)i*(FT)i*(iT)i*(iF)i*(ii)最右推导:EETET*FET*iEF*iEi*iTi*iFi*iii*iETF*TF*FF*(E)F*(ET)F*(EF)F*(Ei)F*(Ti)F*(Fi)F*(ii)i*(ii)语法树:/********************************.精品文档EEEE+TE+TE-TE+TFTT*FE-TFTFiFFiTFiFiiiFiiii+i+ii-i-ii+i*i*****************/P36-9句子iiiei 有两个语法树:S iSeS iSei iiSei iiieiS iS iiSeS iiSei iiieiP36-10/**************TS|TT(S)|()***************/P36-11/***************L1:SACAaAb|abCcC|L2:SABAaA|BbBc|bcL3:.精品文档ABAaAb|BaBb|L4:A|BA0A1|B1B0|A***************/第三章****题参考答案P64–7(1)1(01|)*101X Y01101X12345Y1确定化:01{X}φ{1,2,3}φφφ{1,2,3}{2,3}{2,3,4}{2,3}{2,3}{2,3,4}{2,3,4}{2,3,5}{2,3,4}{2,3,5}{2,3}{2,3,4,Y}{2,3,4,Y}{2,3,5}{2,3,4,}010230001********** 1最小化:.精品文档{0,1,2,3,4,5},{6}{0,1,2,3,4,5}0{1,3,5}{0,1,2,3,4,5}1{1,2,4,6}{0,1,2,3,4},{5},{6}{0,1,2,3,4}0{1,3,5}{0,1,2,3},{4},{5},{6}{0,1,2,3}0{1,3}{0,1,2,3}1{12,,4}{0,1},{2,3}{4},{5},{6}{0,1}0{1}{0,1}1{1,2}{2,3}0{3}{2,3}1{4}{0},{1},{2,3},{4},{5},{6}010 20 0 1 0011345011 1P64–8(1)(1|0)*01(2)(1|2|3|4|5|6|7|8|9)(0|1|2|3|4|5|6|7|8|9)*(0|5)|(0|5)(3)0*1(0|10*1)*|1*0(0|10*1)*P64–12(a)aa,b0 1a确定化:ab{0}{0,1}{1}{0,1}{0,1}{1}{1}{0} φ φ给状态编号:ab0**********aa01abbbb23a最小化:{0,1},{2,3}{0,1}a{1}{0,1}b{2}{2,3}a{0,3}{2,3}b{3}{0,1},{2},{3}aabb012ab(b)bba023abaa bb1a5a4a已经确定化了,:{{0,1},{2,3,4,5}}{0,1}a{1}{0,1}b{2,4}{2,3,4,5}a{1,3,0,5}{2,3,4,5}b{2,3,4,5}{2,4}a{1,0}{2,4}b{3,5}{3,5}a{3,5}{3,5}b{2,4}{{0,1},{2,4},{3,5}}{0,1}a{1}{0,1}b{2,4}{2,4}a{1,0}{2,4}b{3,5}{3,5}a{3,5}{3,5}b{2,4}b b a0 1 2a baP64–1401010(2):X(|)*010Y201X 1 Y0确定化:0 1{X,1,Y} {1,Y} {2}.精品文档{1,Y}{1,Y}{2}{2}{1,Y}φφφφ给状态编号:01012112213333000110111230最小化:{0,1},{2,3}{0,1}0{1}{0,1}1{2}{2,3}0{1,3}{2,3}1{3}{0,1},{2},{3}011101300第四章P81–1按照T,S的顺序消除左递归G(S)a|^|(T)TSTT,ST|递归子程序:procedureS;beginifsym='a'orsym='^'thenabvanceelseifsym='('.精品文档thenbeginadvance;T;ifsym=')'thenadvance;elseerror;endelseerrorend;procedureT;beginS;Tend;procedure T;beginifsym=','thenbeginadvance;S;Tendend;其中:sym:是输入串指针 IP所指的符号advance:是把IP调至下一个输入符号error: 是出错诊察程序(2)FIRST(S)={a,^,(}FIRST(T)={a,^,(}FIRST(T)={,, }FOLLOW(S)={),,,#}FOLLOW(T)={)}FOLLOW(T)={)}预测分析表a ^ ( ) , #S S a S ^ S (T)T T ST T ST T STT T T ,ST是LL(1)文法P81–2文法:.精品文档E TEE E|T FTT|FPFF *F|P (E)|a|b|^(1)FIRST(E)={(,a,b,^}FIRST(E')={+, ε}FIRST(T)={(,a,b,^}FIRST(T')={(,a,b,^, ε}FIRST(F)={(,a,b,^}FIRST(F')={*, ε}FIRST(P)={(,a,b,^}FOLLOW(E)={#,)}FOLLOW(E')={#,)}FOLLOW(T)={+,),#}FOLLOW(T')={+,),#}FOLLOW(F)={(,a,b,^,+,),#}FOLLOW(F')={(,a,b,^,+,),#}FOLLOW(P)={*,(,a,b,^,+,),#}(2)考虑下列产生式 :E E|T T|F *F|P (E)|^|a|bFIRST(+E)∩FIRST(ε)={+}∩{ε}=φFIRST(+E)∩FOLLOW(E')={+}∩{#,)}= φFIRST(T)∩FIRST(ε)={(,a,b,^} ∩{ε}=φFIRST(T)∩FOLLOW(T')={(,a,b,^} ∩{+,),#}= φFIRST(*F') ∩FIRST(ε)={*} ∩{ε}=φFIRST(*F') ∩FOLLOW(F')={*}∩{(,a,b,^,+,),#}= φFIRST((E))∩FIRST(a) ∩FIRST(b) ∩FIRST(^)=φ所以,该文法式 LL(1)文法.(3)+ * ( ) a b ^ #E E TE' E TE'E TE'E TE'E' E E E ET T FT T FT T FT T FTT' T T T T T T T T T T T.