使用排序后的内建函数对python字典的值进行排序(sorting values of python dict using sorted builtin function)

 你已经调用了makeCounter()一次。 这将创建您的新关闭，并将其分配给counter1 。 这个闭包关闭在可变变量上，并且只要闭包存在就会保持捕获。 
 
 调用counter1()将执行它，但它保留相同的捕获n ，并对其进行变异。 这个特殊的“加法器”总是会捕获这个相同的n ，只要它存在.. 
 
 为了获得你所建议的行为，你需要制作新的关闭来捕获n新实例： 
 
let counter1 = makeCounter()

counter1() // returns 1
counter1() // returns 2
counter1() // returns 3

var counter2 = makeCounter()
counter2() // returns 1
counter2 = makeCounter()
counter2() // returns 1
counter2 = makeCounter()
counter2() // returns 1
 
 现在counter1和counter2都有自己独立的n实例。 

You've called makeCounter() once. That creates your new closure, and assigns it to counter1. This closure closes over the mutable var n, and will remain captured as long as this closure exists.
 
Calling counter1() will execute it, but it retains the same captured n, and mutates it. This particular "adder" will ALWAYS capture this same n, so long as it exists..
 
To get the behavior you're suggesting, you need to make new closures which capture new instances of n:
 
let counter1 = makeCounter()

counter1() // returns 1
counter1() // returns 2
counter1() // returns 3

var counter2 = makeCounter()
counter2() // returns 1
counter2 = makeCounter()
counter2() // returns 1
counter2 = makeCounter()
counter2() // returns 1
 
Now both counter1 and counter2 each have their own separate instances of n.