python - "list index out of range"异常(Python3)

Question

当我检查列表 a 的长度时，我不断收到列表索引超出范围异常。第二个 if 语句的 if 或 elif 部分会弹出错误，具体取决于用户输入的内容。我知道当用户输入被分割时，列表会被正确创建，因为我将其打印出来......所以我有点迷失为什么我会收到这个错误。

if __name__ == '__main__':
            for line in sys.stdin:
                    s = line.strip()
                    if not s: break
                    if (str(s) is "quit") == True: quit()
                    elif (str(s) is "quit") == False:
                            a = s.split()
                            print(a)
                            if (len(a) == 2) == True: first(a)
                            elif (len(a) == 3) == True: first(a)
                            else: print("Invalid Input. Please Re-enter.")

第一个方法是:(它在 if 语句中调用的方法只是打印当前的内容)

def first(self, a = list()):
            word = a[0]

            if word is ls:
                    ls(a[1])           
            elif word is format:
                    form(a[1])        # EDIT: was format
            elif word is reconnect:
                    reconnect(a[1])
            elif word is mkfile:
                    mkfile(a[1])
            elif word is mkdir:
                    mkdir(a[1])
            elif word is append:
                    append(a[1], a[2])                               
            elif word is delfile:
                    delfile(a[1])
            elif word is deldir:
                    deldir(a[1])
            else:
                    print("Invalid Prompt. Please Re-enter.")

其他方法:

    def reconnect(one = ""):
            print("Reconnect")

    def ls(one = ""):
            print("list")

    def mkfile(one = ""):
            print("make file")

    def mkdir(one = ""):
            print("make drive")

    def append(one = "", two = ""):
            print("append")

    def form(one = ""):
            print("format")

    def delfile(one = ""):
            print("delete file")

    def deldir(one = ""):
            print("delete directory")

    def quit():
            print("quit")
            sys.exit(0)

Answer 1

问题似乎出在 first() 的定义上。您将其作为函数调用:

if (len(a) == 2) == True: first(a)
elif (len(a) == 3) == True: first(a)

但是你将它定义为一个方法:

def first(self, a = list()):

命令和参数的数组被放入 self 中，并且 a 始终是一个空列表，您尝试对其进行索引并失败。另外，除非您确定自己在做什么，否则不应使用 list() 这样的可变类型作为默认值。我简单地建议:

def first(a):

就您的 __main__ 代码而言，简化:

if __name__ == '__main__':

    for line in sys.stdin:
        string = line.strip()

        if not string:
            break

        if string == "quit":
            quit()

        tokens = string.split()

        length = len(tokens)

        if 2 <= length <= 3:
            first(tokens)
        else:
            print("Invalid Input. Please Re-enter.")