为什么不能在函数下覆盖变量？

Question

a = int(input("How many pizza would you like to order? "))                   #line 1

def count_ordered_pizza():
    if a==0:
        print("Okay see u next time!")
    elif a==1:
        print(str(a)+ " pizza ordered.")
    elif a>1:
        print(str(a)+ " pizzas ordered.")
    else:
        print("Sorry that's impossible")

def reorder():
    user_said= str(input('Do you wish to order again? State Y/N. '))
    while user_said != "N":
        a = int(input("How many pizza would you like to order? "))          #line 16
        count_ordered_pizza()
        reorder()
    print('Bye!')

如果我执行此操作，reorder() 函数（第 16 行）中的 a 不会替换 count_ordered_pizza() 函数（第 1 行）中的 a。这是为什么？

我认为它可以替换它们，因为变量已被覆盖，但如果我这样做：

def reorder():
    user_said= str(input('Do you wish to order again? State Y/N. '))
    while user_said != "N":
        a = int(input("How many pizza would you like to order? "))          #line 16
        if a==0: 
            print("Okay see u next time!")
        elif a==1:
            print(str(a)+ " pizza ordered.")
        elif a>1:
            print(str(a)+ " pizzas ordered.")
        else:
            print("Sorry that's impossible")

        reorder()
    print('Bye!')

它显示了正确的值。有人可以帮我吗？提前致谢！

Answer 1

如果在函数 global explanation

中使用 global a 表达式，则可以更改全局变量值

a = int(input("How many pizza would you like to order? "))                   #line 1

def count_ordered_pizza():
    if a==0:
        print("Okay see u next time!")
    elif a==1:
        print(str(a)+ " pizza ordered.")
    elif a>1:
        print(str(a)+ " pizzas ordered.")
    else:
        print("Sorry that's impossible")

def reorder():
    user_said= str(input('Do you wish to order again? State Y/N. '))
    while user_said != "N":
        global a
        a = int(input("How many pizza would you like to order? "))          #line 16
        count_ordered_pizza()
        reorder()
    print('Bye!')

为什么不能在函数下覆盖变量？

Why can't a variable be overwritten under a function?

python

variables

function

while-loop