消除左递归--作业10

1.将以下文法消除左递归,并分析符号串 i*i+i  

并分别求FIRST集、FOLLOW集,和SELECT集

     E -> E+T | T

     T -> T*F | F

     F -> (E) | i

解:  E -> TE'

    E' -> +TE' | ε

    T -> FT'

    T' -> *FT' | ε

    F -> (E) | i

FRIST集: FRIST( TE' ) -> { ( , i }

      FRIST( +TE' ) -> { + }

      FRIST( ε ) -> { ε }

      FRIST( FT' ) -> { F }

      FRIST( *FT' ) -> { * }

      FRIST( ε ) -> { ε }

      FRIST( (E) ) -> { ( }

      FRIST( i ) -> { i }

FOLLOW集:   FOLLOW( E ) -> { ) }

        FOLLOW( E' ) -> { # }

        FOLLOW( F ) -> { * }

        FOLLOW( T ) -> { + }

        FOLLOW( T' ) -> { * }

SELECT集:  SELECT( E -> TE' ) -> { ( , i }

        SELECT( E' -> +TE' ) -> { + }

        SELECT( E' -> ε ) -> { # }

        SELECT( T -> FT' ) -> { F }

        SELECT( T' -> *FT' ) -> { * }

        SELECT( T' -> ε ) -> { * }

        SELECT( F -> (E) ) -> { ( }

        SELECT( F -> i ) -> { i }

分析符号串  i * i + i

 

2.P101练习7(2)(3)文法改写,并分别求FIRST集、FOLLOW集,和SELECT集

(2)     A -> aA'

    A' -> aA'Be | ε

    B -> dB'

    B' -> bB' | ε

FIRST集:  FIRST( aA' ) -> { a }

         FIRST( aA'Be ) -> { a}

              FIRST( ε ) -> { ε }

         FIRST( dB' ) -> { d }

         FIRST( bB' ) -> { b }

         FIRST( ε ) -> { ε }

FOLLOW集:  FOLLOW( A ) -> { # }

         FOLLOW( A' ) -> { e }

         FOLLOW( B ) -> { e }

         FOLLOW( B' ) -> { b , # }

SELECT集:  SELECT( A -> aA' ) -> { a }

        SELECT( A' -> aA'Be ) -> { a }

        SELECT( A' -> ε ) -> { e }

        SELECT( B -> dB' ) -> { d }

        SELECT( B' -> bB' ) -> { b }

        SELECT( B' -> ε ) -> { b , # }

(3)  S -> Aa | b -> Saba | b

    S -> bS'

    S' -> abaS'

FIRST集:  FIRST( bS' ) -> { b }

         FIRST( abaS' ) -> { a }

FOLLOW集:  FOLLOW( S ) -> { # }

         FOLLOW( S' ) -> { # }

SELECT集:  SELECT( S -> bS' ) -> { b }

        SELECT( S' -> abaS' ) -> { a }

 

课堂练习:

求以下文法的FIRST集、FOLLOW集和SELECT集。

S->Ap
A->a | ε
A->cA
A->aA

解:FIRST( Ap ) = { a,c,p }

  FIRST( a ) = { a }

  FIRST( ε ) = { ε }

  FIRST( cA ) = { c }

  FIRST( aA ) = { a }

 

  FOLLOW( S ) = { # }

  FOLLOW( A ) = { p }

  

  SELECT( S -> Ap ) -> { a,c,p }

  SELECT( S -> a ) -> { a }

  SELECT( S -> ε ) -> { p }

  SELECT( S -> cA ) -> { c }

  SELECT( S -> aA ) -> { a }

 

S->Ap
S->Bq
A->a
A->cA
B->b
B->dB

解:FIRST( Ap ) = { a,c }

  FIRST( Bq ) = { b,d }

  FIRST( a ) = { a }

  FIRST( cA ) = { c }

  FIRST( b ) = { b }

  FIRST( dB ) = { d }

 

  FOLLOW( S ) = { # }

  FOLLOW( A ) = { p }

  FOLLOW( b ) = { q }

  

  SELECT( S -> Ap ) -> { a,c }

  SELECT( S -> Bq ) -> { b,d }

  SELECT( S -> a ) -> { a }

  SELECT( S -> cA ) -> { c }

  SELECT( S -> b ) -> { b }

  SELECT( S -> dB ) -> { d }

posted @ 2019-11-14 21:53  ·无语·  阅读(225)  评论(0编辑  收藏  举报