Python 如何将列表从一个函数传递到另一个函数?
问题:玩弄在函数之间交换变量的想法 我正在执行以下代码:Python 如何将列表从一个函数传递到另一个函数?,python,python-3.x,list,function,Python,Python 3.x,List,Function,问题:玩弄在函数之间交换变量的想法 我正在执行以下代码: def benefits(): list = ["Beautiful", "Explicit", "Simple", "Readability","Easy to share"] return list def statement(benefit): print("The benifit is " + benefit ) def benefits_of_functions():
def benefits():
list = ["Beautiful", "Explicit", "Simple", "Readability","Easy to share"]
return list
def statement(benefit):
print("The benifit is " + benefit )
def benefits_of_functions():
benefits_list = benefits()
for benefit in benefits_list:
print(statement(benefit))
benefits_of_functions()
我得到一个错误:
The benifit is Beautiful
None
The benifit is Explicit
None
The benifit is Simple
None
The benifit is Readability
None
The benifit is Easy to share
None
我不能理解“没有”。您能帮我弄清楚为什么会出现在输出中吗?在
语句中返回而不是打印函数:
def benefits():
list = ["Beautiful", "Explicit", "Simple", "Readability","Easy to share"]
return list
def statement(benefit):
return "The benifit is " + benefit
def benefits_of_functions():
benefits_list = benefits()
for benefit in benefits_list:
print(statement(benefit))
benefits_of_functions()
因为您的语句
函数只打印一些内容,不返回任何内容,所以它隐式返回None
。在函数的优点
函数中,循环列表,然后打印将列表中的项目传递到语句的结果,该语句将打印消息,但您正在打印该结果,即无
def语句
应该只返回“好处是”+好处
,而不是打印它。注意,重要的是要理解,不是将变量传递到函数中,而是传递对象。