由于BNF语法,生成AST(抽象语法树)时出错

时间:2019-04-06 15:58:54

标签: python lambda alpha lambda-calculus bnf

所以,我有一个语法,即:

    termo : PESQ termo termo PDIR

     | PESQ LAMBDA listav PONTO termo PDIR

     | VAR

    listav : VAR listav
      | VAR

其中termo = term,PESQ =(,PDIR =),LAMBDA =!,PONTO =。,VAR = [a-z],listav =变量列表。

问题是,当我尝试生成我输入的某些表达式的抽象语法树时,例如!x.x,它给了我一个错误:yacc:第1行的语法错误,token = LAMBDA。

我认为这里的问题是我定义语法的方式(并将非终结符插入堆栈数组。我将代码留在底部。也许是关于stack.append(p [x ])?我不确定应该将哪些非终结符附加到数组,因为我有多个生成。

import sys
from tree import *

stack = []

tokens = [

    'LAMBDA',
    'PONTO',
    'VAR',
    'PDIR',
    'PESQ'
]


t_LAMBDA = r'\!'
t_PONTO = r'\.'
t_VAR = r'[a-z]'
t_PDIR = r'\)'
t_PESQ = r'\('

t_ignore = r' '


def t_error(t):
    print("Illegal Character!")
    t.lexer.skip(1)

import ply.lex as lex
lexer = lex.lex()

def p_termo(p):
    '''
    termo : PESQ termo termo PDIR
          | PESQ LAMBDA listav PONTO termo PDIR
          | VAR
    '''

    if len(p) == 7:
        stack.append(p[2])
        stack.append(p[3])
        stack.append(p[4])
    elif len(p) == 2:
        stack.append(p[1])


def p_listav(p):
    '''
    listav : VAR listav
           | VAR
    '''

    if len(p) == 3:
        stack.append(p[1])
    elif len(p) == 2:
        stack.append(p[1])

import ply.yacc as yacc
parser = yacc.yacc()




while True:
    try:
        print('')
        print('Input: ')
        inp = input('')
    except EOFError:
        break
    stack = []
    p = parser.parse(inp)


    z = 0
    while z < len(stack):
        if z == 0:
            if len(stack) == 1:
                a = Tree(stack[z])
                z += 1
            elif stack[z+1] != '!':
                auxL = Tree(stack[z + 1])
                auxR = Tree(stack[z])
                a = Tree('@', auxL, auxR)
                z += 2
            else:
                a = Tree(stack[z])
                z += 1

        else:
            if stack[z] == '!' and stack[z-2] == '.' and z > 1:
                auxR = Tree(stack[z + 1])
                b = Tree('!', m, auxR)
                a = Tree('@', a, b)
                z += 3
            elif stack[z] == '!':
                auxL = Tree(stack[z+1])
                a = Tree('!', auxL, a)
                z += 3
            elif stack[z] != '!' and stack[z-1] == '.' and z != len(stack) - 1:
                if stack[z+1] == '!':
                    m = Tree(stack[z])
                    z += 1
                else:
                    auxR = Tree(stack[z])
                    a = Tree('@', a, auxR)
                    z += 1
            else:
                auxR = Tree(stack[z])
                a = Tree('@', a, auxR)
                z += 1



        arv = Tree.print_tree(a)
        print('###########################')
        print('Arvore: ')
        print(arv)
        print('###########################')

0 个答案:

没有答案