Question

这个函数不会从脚本中调用变量吗？这与Zed Shaw所说的相反......

我已将评论放在自己身上，我的问题是：

Zed Shaw说：

函数中的变量未连接到脚本中的变量

但是从我看到的情况来看，函数在缩进时位于顶部，然后当缩进开始创建变量然后链接到函数时。

这个函数不是从脚本调用变量吗？

我无法找到合适的答案：有人可以详细说明这段脚本并告诉我我是否错误地查看了它？

# This is the function with declared variables inside
def cheese_and_crackers(cheese_count, boxes_of_crackers):
    print  "You have %d cheeses!" % cheese_count
    print "You have %d boxes of crackers!"  % boxes_of_crackers
    print "Man thats enough for a party!"
    print "Get a Blanket. \n"

# This declares the amounts in the functions variables
print "We can just give the function numbers directly:"
cheese_and_crackers (20, 30)

# This variable sets the amounts
print "OR, we can use variables from our script:"
amount_of_cheese = 10
amount_of_crackers = 50
#This variable combines the two above and stores it in a single variable for the function to call
cheese_and_crackers(amount_of_cheese, amount_of_crackers)

# This uses the function, but has predefined variables with maths
print "We can even do math inside too:"
cheese_and_crackers(10 + 20, 5 + 6)
#This combines variables with maths
print "And we can combine the two, variables and math:"
cheese_and_crackers (amount_of_cheese + 100, amount_of_crackers + 1000)

Answer 1

变量不会在任何时候连接到函数参数，除了它们作为参数传递给函数的事实。调用foo(bar)不会将bar连接到foo()，它只会将bar的值作为函数的第一个参数传递。如果该参数也恰好被称为“bar”，那么这就是巧合。

Python函数和变量 - 学习Python ex19的难点

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