How does a simple calculator with parentheses work?

后端 未结 2 1212
我寻月下人不归
我寻月下人不归 2020-12-05 05:05

I want to learn how calculators work. For example, say we have inputs in infix notation like this:

1 + 2 x 10 - 2

The parser would have to respect common rul

相关标签:
2条回答
  • 2020-12-05 05:45

    Try looking at Antlr. It is what I used to build a custom compiler/parser... and could easily relate to a calculator which would be a very simple thing to create.

    0 讨论(0)
  • 2020-12-05 05:49

    One way to do evaluate an expression is with a recursive descent parser. http://en.wikipedia.org/wiki/Recursive_descent_parser

    Here's an example grammar in BNF form: http://en.wikipedia.org/wiki/Backus-Naur_form

    Expr ::= Term ('+' Term | '-' Term)*
    Term ::= Factor ('*' Factor | '/' Factor)*
    
    Factor ::= ['-'] (Number | '(' Expr ')')
    
    Number ::= Digit+
    

    Here * means the preceding element is repeated zero or more times, + means one or more repeats, square brackets means optional.

    The grammar ensures that the elements of highest precedence are collected together first, or in this case, evaluated first. As you visit each node in the grammar, instead of building an abstract syntax tree, you evaluate the current node and return the value.

    Example code (not perfect but should give you an idea of how to map BNF to code):

    def parse_expr():
      term = parse_term()
      while 1:
        if match('+'):
          term = term + parse_term()
        elif match('-'):
          term = term - parse_term()
        else: return term
    
    def parse_term():
      factor = parse_factor()
      while 1:
        if match('*'):
          factor = factor * parse_factor()
        elif match('/'):
          factor = factor / parse_factor()
        else: return factor
    
    def parse_factor():
      if match('-'):
        negate = -1
      else: negate = 1
      if peek_digit():
        return negate * parse_number()
      if match('('):
        expr = parse_expr()
        if not match(')'): error...
        return negate * expr
      error...
    
    def parse_number():
      num = 0
      while peek_digit():
        num = num * 10 + read_digit()
      return num
    

    To show how your example of 1 + 2 * 10 - 2 would evaluate:

    call parse_expr                              stream is 1 + 2 * 10 - 2
      call parse term
        call parse factor
          call parse number which returns 1      stream is now + 2 * 10 - 2
        match '+'                                stream is now 2 * 10 - 2
        call parse factor
          call parse number which returns 2      stream is now * 10 - 2
          match '*'                              stream is now 10 - 2
          call parse number which returns 10     stream is now - 2
          computes 2 * 10, return 20
        compute 1 + 20 -> 21
        match '-'                                stream is now 2
        call parse factor
          call parse number which returns 2      stream is empty
        compute 21 - 2, return 19
      return 19
    
    0 讨论(0)
提交回复
热议问题