Question

我正在为经验公式编写语法控制器。它似乎工作得相当好，但有一个我似乎无法解决的错误。

以下是代码：

import string
from linkedQFile import LinkedQ

ATOMER = ["H","He","Li","Be","B","C","N","O","F","Ne","Na","Mg","Al","Si","P","S","Cl","Ar"]

class Formelfel(Exception):
    pass

def readFormel():
    if q.peek() == None:
        print("Formel saknas. ")
    readMol()
    print("Formeln är syntaktiskt korrekt")
    #newline check?

def readMol():
    if q.peek() == None:
        return
    readGroup()
    readMol()
    return

def readGroup():
    if q.peek() == "(":
        print(q.get(), end="")
        if q.peek() in string.ascii_lowercase or q.peek() in string.ascii_uppercase:
            readMol()
        else:
            raise Formelfel
    if q.peek() == ")":
        print(q.get(), end="")
        readNum()
        return
    readAtom()
    try: 
        readNum()
    except Formelfel:
        pass
    return

def readAtom():
    X = readLetter()
    try:
        x = readletter() 
    except Formelfel:
        x = ""
    atom = X+x
    if atom in ATOMER:
        return
    rest=""
    while not q.isEmpty():
        rest += print(q.get(), end="")
    raise Formelfel("Okänd atom vid radslutet "+rest)

def readNum():
    try:
        if q.peek() != None:
            print(q.peek())
            while not q.peek() in string.ascii_uppercase and not q.peek() in string.ascii_lowercase and not q.peek() == "(" and not q.peek()==")": # If q.peek() is a number
                if q.peek() == None:
                    return
                int(q.get())    
        return
    except:
        raise Formelfel

def readLetter():
    if q.peek() in string.ascii_uppercase:
        x =q.get()
        print(x,end="")
        return x
    raise Formelfel

def readletter():
    if q.peek() == None:
        raise Formelfel
    if q.peek() in string.ascii_lowercase:
        x =q.get()
        print(x,end= "")
        return x
    raise Formelfel

formel= "Si(C3(COOH)2)4(H2O)7"
q = LinkedQ()
for symbol in formel:
    q.put(symbol)

readFormel()

我收到以下输出和错误：

Si(
(C3
(CO
OO
OH
H)
)2
)4
(H2
O)
)7
Traceback (most recent call last):
  File "formelkoll.py", line 59, in readNum
    while not q.peek() in string.ascii_uppercase and not q.peek() in string.ascii_lowercase and not q.peek() == "(" and not q.peek()==")": # If q.peek() is a number
TypeError: 'in <string>' requires string as left operand, not NoneType

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "formelkoll.py", line 88, in <module>
    readFormel()
  File "formelkoll.py", line 12, in readFormel
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 19, in readMol
    readGroup()
  File "formelkoll.py", line 27, in readGroup
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 19, in readMol
    readGroup()
  File "formelkoll.py", line 27, in readGroup
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 19, in readMol
    readGroup()
  File "formelkoll.py", line 27, in readGroup
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 20, in readMol
    readMol()
  File "formelkoll.py", line 19, in readMol
    readGroup()
  File "formelkoll.py", line 32, in readGroup
    readNum()
  File "formelkoll.py", line 65, in readNum
    raise Formelfel
__main__.Formelfel

我不明白为什么会出现这种错误，因为我有一个if语句：

if q.peek() != None:

发生错误的行之前。换句话说，为什么 None 允许通过这个if语句？

Answer 1

我明白了。正如评论者所说，调用q.get（），它会改变q.peek（）的值。两行的切换解决了这个问题。

def readNum():
    try:
        if q.peek() != None: #Vrf får vi error NoneType trots detta??
            print(q.peek())
            while not q.peek() in string.ascii_uppercase and not q.peek() in string.ascii_lowercase and not q.peek() == "(" and not q.peek()==")": 
                int(q.get())    
                if q.peek() == None: 
                    return

        return
    except:
        raise Formelfel

if语句没有捕获无

1 个答案: