函数 - 关于python闭包的一些疑问

Question

def count():
    fs = []
    for i in range(1, 4):
        def f(j):
            def g():
                return j*j
            return g    
        r = f(i)    
        fs.append(r)    #fs作为列表，为什么可以append(r)？(而r是f所返回的函数)
    return fs           #个人的想法应该是fs.append(r())

f1, f2, f3 = count()
print f1(), f2(), f3()

想弄明白的是：这里传入append的r是以什么形式传入的呢
谢谢各位指教:D

Answer 1

is passed in in the form of a function definition, similar to delayed calling, which is not obvious in your example,

def count():
    def r():
        return "hello world"
    return r          
x =  count()  
print x #<function r at 0x7fc562978668>
print x() # hello world

You can think of r in return r in the function as a variable, but this variable saves the definition of the function.
The actual call to the function is through x() later

Answer 2

r is an object, and the type of this object is a function.

In other words, r is an object of type function, like

• 1 is an object of type integer

• 'hi' is the same as an object of type string

Answer 3

If you change it to fs.append(r()), it is also possible, but the meaning of the function is different. After executing the count() function, it will return the result [1, 4, 9], and you can no longer use print f1(), f2(), f3() to print out the results because: fs.append(r())也是可以的，但是函数意义就不同了，执行count()函数后其会返回结果[1, 4, 9]，而不能再使用print f1(), f2(), f3()打印出结果，原因在于：

r=f(i)=g

注意这里是 g 不是 g(),所以执行count()返回的其实是[g1, g2, g3]一个包含三个闭包函数的列表，每个闭包函数绑定的自由变量不同（分别是1， 2， 3），因此将这个列表解包给f1, f2, f3后，执行print f1(), f2(), f3()相当于执行三个函数后并打印出结果，因此最终输出1, 4, 9 rrreee

Note that this is g not g(), so what is returned by executing count() is actually [g1, g2, g3] A list containing three closure functions. Each closure function binds different free variables (respectively 1, 2, 3), so unpack this list to f1, f2, f3 , executing print f1(), f2(), f3() is equivalent to executing three functions and printing out the results, so the final output is 1, 4, 9< /code>🎜

Answer 4

f1 = fs[0] = r = gf1 = fs[0] = r = g

所以才能

f1()

That’s why 🎜 🎜f1()🎜