陈火旺编译原理(第三版)课后习题答案 下载本文

26 #((T) ,a)# 进 27 #(S ,a)# 归 28 #(T ,a)# 归 29 #(T, a)# 进 30 #(T,a )# 进 31 #(T,S )# 归 32 #(T )# 归 33 #(T) # 进

34 #S # 归

P133–3

(1)

FIRSTVT(S)={a,^,(} FIRSTVT(T)={,,a,^,(} LASTVT(S)={a,^,)} LASTVT(T)={,,a,^,)} (2) a ^ ( ) , a > > ^ > > ( < < < = < ) > > , < < < > > G6是算符文法,并且是算符优先文法

(3)优先函数 a ^ ( ) , f 4 4 2 4 4 g 5 5 5 2 3

fa f^ f(

f) f,

ga g^ g( g) g,(4) 栈 输入字符串 # (a,(a,a))# #( a, (a,a))# #(a , (a,a))# #(t

, (a,a))#

. . .

动作 预备进 进 归

#(t, #(t,( #(t,(a #(t,(t #(t,(t, #(t,(t,a #(t,(t,s #(t,(t #(t,(t) #(t,s #(t #(t ) # s success

(a,a))# a,a))# ,a))# ,a))# a))# ))# ))# ))# )# )# )# # # 进 进 进 归 归

进 进 归 进 归 归

归 进

P134–5

(1)

0.S???S 1.S??S? 2.S??AS 3.S?A?S 4.S?AS? 5.S??b 6.S?b? 7.A??SA 8.A?S?A 9.A?SA? 10.A??a 11.A?a? (2)

1

S A ? 8 7 9 S ? ? ? ? a 10 0

? ? ? A S 3 2 ? ?

d 5

确定化: {0,2,5,7,10} {1,2,5,7,8,10S {1,2,5,7,8,10} {2,5,7,8,10} A {2,3,5,7,10} {2,3,5,7,9,1011 4 6 a {11} {11} b {6} {6} . . .

} {2,3,5,7,10} {2,5,7,8,10} {2,3,5,7,9,10} {2,4,5,7,8,10} {11} {6} {2,4,5,7,8,10} {2,5,7,8,10} {2,4,5,7,8,10} {2,5,7,8,10} φ φ } {2,3,5,7,10} {2,3,5,7,9,10} {2,3,5,7,10} {2,3,5,7,9,10} φ φ {11} {11} {11} {11} φ φ {6} {6} {6} {6} φ φ

A S

5:A?S?A 6:A?SA? 3:S??S? S??AS S?A?S S A a A?S?A S??AS S??b

A??SA A??SA S??b b A??a A??a A??SA

S??ASA??a

S??b

S a A S b S A b a A

4:S?A?S 7:S?AS? 0:S???S

S??AS S??AS A?S?A A S S??AS S??b S??b

A??SA A??SA S??b b A??a A??a A??SA A??a

a a b b a

1:A?a? 2:S?b?

DFA

构造LR(0)项目集规范族也可以用GO函数来计算得到。所得到的项目集规范族与上图中的项目集一样:

I0={S???S,S??AS,S??b,A??SA,A??a} GO(I0,a)={ A?a?}=I1 GO(I0,b)={ S?b?}=I2

GO(I0,S)={ S??S?,A?S?A,A??SA,A??a,S??AS,S??b}=I3

. . .

GO(I0,A)={ S?A?S,S??AS,S??b,A??SA,A??a}=I4 GO(I3,a)={ A?a?}=I1 GO(I3,b)={ S?b?}=I2

GO(I3,S)={ A?S?A,S??AS,S??b,A??SA,A??a}=I5

GO(I3,A)={ A?SA?,S?A?S,S??AS,S??b,A??SA,A??a}=I6 GO(I4,a)={ A?a?}=I1 GO(I4,b)={ S?b?}=I2

GO(I4,S)={ S?AS?,A?S?A,S??AS,S??b,A??SA,A??a}=I7 GO(I4,A)={ S?A?S,S??AS,S??b,A??SA,A??a}=I4 GO(I5,a)={ A?a?}=I1 GO(I5,b)={ S?b?}=I2

GO(I5,S)={ A?S?A,S??AS,S??b,A??SA,A??a}=I5

GO(I5,A)={ A?SA?,S?A?S,S??AS,S??b,A??SA,A??a}=I6 GO(I6,a)={ A?a?}=I1 GO(I6,b)={ S?b?}=I2

GO(I6,S)={ S?AS?,A?S?A,S??AS,S??b,A??SA,A??a}=I7 GO(I6,A)={ S?A?S,S??AS,S??b,A??SA,A??a}=I4 GO(I7,a)={ A?a?}=I1 GO(I7,b)={ S?b?}=I2

GO(I7,S)={ A?S?A,S??AS,S??b,A??SA,A??a}=I5

GO(I7,A)={ A?SA?,S?A?S,S??AS,S??b,A??SA,A??a}=I6 项目集规范族为C={I1,I2,I3,I4,I5,I6,I7}

(3)不是SLR文法

状态3,6,7有移进归约冲突

状态3:FOLLOW(S’)={#}不包含a,b

状态6:FOLLOW(S)={#,a,b}包含a,b,;移进归约冲突无法消解 状态7:FOLLOW(A)={a,b}包含a,b;移进归约冲突消解 所以不是SLR文法。

(4) 构造例如LR(1)项目集规范族 见下图: 对于状态5,因为包含项目[A?AS? a/b],所以遇到搜索符号a或b时,应该用A?AS归约。又因为状态5包含项目[A??a a/b],所以遇到搜索符号a时,应该移进。因此存在“移进-归约”矛盾,所以这个文法不是LR(1)文法。

. . .