Question

对于类，我们得到了一个充满语法错误的源代码。我修复了一些更明显的语法问题。这是迄今为止的代码：

def findNumInList():
    for i in (myList):
        if(myList[i] == num):
            print("Found number %s" %num)

def main():
    myList = [1,25,7,99,12]

    #Gets number from user, and appends it to the existing list
    num = int(input("Enter a number to be added to the end of the list: "))
    myList.append(num)

    #Checks to see if number was successfully added to the list
    findNumInList()
main()

我还得到的是：

Traceback (most recent call last):
  File "part1.py", line 16, in <module>
    main()
  File "part1.py", line 15, in main
    findNumInList()
  File "part1.py", line 3, in findNumInList
    for i in (myList):
NameError: global name 'myList' is not defined

myList如何定义？

Answer 1

首先调用

main()，并在那里定义列表，但它只存在于该函数的范围内，因此findNumInList函数不知道它。

解决方案是将列表传递给函数：

def findNumInList(myList, num):
    for i in (myList):
        if(myList[i] == num):
            print("Found number %s" %num)

def main():
    myList = [1,25,7,99,12]

    #Gets number from user, and appends it to the existing list
    num = int(input("Enter a number to be added to the end of the list: "))
    myList.append(num)

    #Checks to see if number was successfully added to the list
    findNumInList(myList, num)
main()

编辑：同样适用于num变量

Answer 2

考虑阅读Python中scope的内容：

[...]通常，本地范围引用（文本）当前函数的本地名称。在外部函数中，本地作用域引用与全局作用域相同的名称空间：模块的名称空间。 [...]

您的变量位于main函数的范围内，该函数是本地范围。您无法访问本地范围之间的变量。作为Tim Castelijns showed in his answer，一种可能的解决方案是将列表作为参数传递。

搜索语法错误

2 个答案: