来自BNF的递归下降解析器伪代码

Question

Ok, I thought there would be enough CS majors on here to check my pseudo code for my recursive descent parser. I developed it from this BNF

EXP ::= EXP + TERM | EXP - TERM | TERM
TERM ::= TERM * FACTOR | TERM/FACTOR | FACTOR
FACTOR ::= (EXP) | DIGIT
DIGIT ::= 0|1|2|3

Here's the pseudo code:

procedure exp()
    term()
    if token == ‘+’
        match(‘+’)
        term()
    elseif token == ‘-‘
        match(‘-‘)
        term()
    else
        break    

procedure term()
    factor()
    if token == ‘*’
        match(‘*’)
        factor()
    elseif token == ‘/’
        match(‘/’)
        factor()
    else
        break

procedure factor()
    if token == ‘(‘
        match(‘(‘)
        exp()
        match(‘)’)
    else
        digit()

procedure digit()
    if token == ‘0’
        match(‘0’)
    elseif token == ‘1’
        match(‘1’)
    elseif token == ‘2’
        match(‘2’)
    else
        match(‘3’)

match(t)
    if token == t
        advancetokenpointer
    else
        error

Is this correct? I was thinking I may need to have a return in each procedure, and I'm not sure of my procedure correctness either. Maybe include end procedure too? Anyways, thanks a lot! :-)

Answer 1

You are half way there. Especifically, you are not accounting for the left-recursive parts of the grammar, as in "EXP ::= EXP...", or "TERM ::= TERM...".

Recursive descent is not well suited for left-recursion anyway, but fortunatelly there are standard transformations you can perform in the grammar that will eliminate this kind of left recursion. As an example, the following grammar:

A ::= A x B | B

could be coded like this:

procedure A()
  B()
  repeat
    if token == 'x'
      match('x')
      B()
    else
      break

Moreover, the code for factor is not following the grammar correctly. Notice that in the first alternative, EXP is called recursivelly (this is a kind of recursion that Recursive Descent has no problem with) and you are calling factor instead. Also, you are matching the right parenthesis as if it was optional, while it is actually required. This same problem exists in the code for DIGIT. If neither 0, 1 or 2 match, 3 must match.