无界法TypeError

Question

我刚刚读到一篇关于在python中实现解析器的文章：http://effbot.org/zone/simple-top-down-parsing.htm

本文描述了代码背后的一般思想：http://mauke.hopto.org/stuff/papers/p41-pratt.pdf

对于用python编写解析器来说，这是相当新的，所以我尝试编写类似于学习练习的东西。然而，当我试图编写类似于本文中发现的代码时，我得到了一个TypeError: unbound method TypeError。这是我第一次遇到这样的错误，我花了一整天的时间来解决这个问题，但是我还没有解决这个问题。下面是一个有此问题的最小代码示例(完整的代码示例)：

import re

class Symbol_base(object):
    """ A base class for all symbols"""
    id = None # node/token type name
    value = None #used by literals
    first = second = third = None #used by tree nodes

    def nud(self):
        """ A default implementation for nud """
        raise SyntaxError("Syntax error (%r)." % self.id)

    def led(self,left):
        """ A default implementation for led """
        raise SyntaxError("Unknown operator (%r)." % self.id)

    def __repr__(self):
        if self.id == "(name)" or self.id == "(literal)":
            return "(%s %s)" % (self.id[1:-1], self.value)
        out = [self.id, self.first, self.second, self.third]
        out = map(str, filter(None,out))
        return "(" + " ".join(out) + ")"


symbol_table = {}
def symbol(id, bindingpower=0):
    """ If a given symbol is found in the symbol_table return it.
        If the symblo cannot be found theni create the appropriate class
        and add that to the symbol_table."""
    try:
        s = symbol_table[id]
    except KeyError:
        class s(Symbol_base):
            pass
        s.__name__ = "symbol:" + id #for debugging purposes
        s.id = id
        s.lbp = bindingpower
        symbol_table[id] = s
    else:
        s.lbp = max(bindingpower,s.lbp)
    return s

def infix(id, bp):
    """ Helper function for defining the symbols for infix operations """
    def infix_led(self, left):
        self.first = left
        self.second = expression(bp)
        return self
    symbol(id, bp).led = infix_led

#define all the symbols
infix("+", 10)
symbol("(literal)").nud = lambda self: self #literal values must return the symbol itself
symbol("(end)")

token_pat = re.compile("\s*(?:(\d+)|(.))")

def tokenize(program):
    for number, operator in token_pat.findall(program):
        if number:
            symbol = symbol_table["(literal)"]
            s = symbol()
            s.value = number
            yield s
        else:
            symbol = symbol_table.get(operator)
            if not symbol:
                raise SyntaxError("Unknown operator")
            yield symbol
    symbol = symbol_table["(end)"]
    yield symbol()

def expression(rbp = 0):
    global token
    t = token
    token = next()
    left = t.nud()
    while rbp < token.lbp:
        t = token
        token = next()
        left = t.led(left)
    return left

def parse(program):
    global token, next
    next = tokenize(program).next
    token = next()
    return expression()

def __main__():
    print parse("1 + 2")

if __name__ == "__main__":
    __main__()

当我尝试用pypy运行它时：

Traceback (most recent call last):
  File "app_main.py", line 72, in run_toplevel
  File "parser_code_issue.py", line 93, in <module>
    __main__()
  File "parser_code_issue.py", line 90, in __main__
    print parse("1 + 2")
  File "parser_code_issue.py", line 87, in parse
    return expression()
  File "parser_code_issue.py", line 81, in expression
    left = t.led(left)
TypeError: unbound method infix_led() must be called with symbol:+ instance as first argument (got symbol:(literal) instance instead)

我猜这是因为我没有为infix操作创建一个实例，但是我并不想在那个时候创建一个实例。有什么方法可以不用创建实例来更改这些方法吗？

任何帮助解释为什么会发生这种情况，以及我能做些什么来修复代码是非常感谢的！

在python 3中，这种行为也会改变吗？

Answer 1

忘记在tokenize()函数中创建符号的实例；如果不是数字，则生成symbol()，而不是symbol。

else:
    symbol = symbol_table.get(operator)
    if not symbol:
        raise SyntaxError("Unknown operator")
    yield symbol()

通过这一项更改，您的代码打印：

(+ (literal 1) (literal 2))

Answer 2

您还没有将新函数绑定到对象的实例。

import types

obj = symbol(id, bp)
obj.led = types.MethodType(infix_led, obj)

见another SO question接受的答案