Python默认参数评估 [英] Python Default Arguments Evaluation

问题描述

我正在阅读python文档版本2.7.10中的python教程，遇到了这样的事情。

I was reading the python tutorial from Python Documentation Release 2.7.10 and I came across something like this.

代码

def fun1(a,L=[]):
    L.append(a)
    return L

print fun1(1)
print fun1(2)
print fun1(3)

def fun2(a,L = None):
    if L is None:
        L=[]
    L.append(a)
    return L

print fun2(1)
print fun2(2)
print fun2(3)

输出

[1]
[1, 2]
[1, 2, 3]
[1]
[2]
[3]

Process finished with exit code 0

如果第一个函数 fun1（）中的 L = [] 仅调用一次， fun1（）的输出就可以了。但是，为什么每次在 fun2（）中都会调用 L = None 。

If the L=[] in the first function fun1() is getting called only once , the output of fun1()is fine. But then why L=None is getting called every time in fun2().

推荐答案

定义函数时，将评估默认参数的值，但但函数主体仅被编译。您可以通过属性检查函数定义的结果。有一个 __ defaults __ 属性，包含默认值，而 __ code __ 属性，包含主体（因此在定义函数时创建）

When you define a function the values of the default arguments get evaluated, but the body of the function only get compiled. You can examine the result of the function definition via attributes. Theres a __defaults__ attribute containing the defaults and __code__ attribute containing the body (so these are created when the function is defined).

第二个示例中发生的事情是 None 确实在定义时得到了评估（其评估结果为 None duh！），但是有条件地将 [] 分配给 L 仅被编译并每次都运行（条件通过）。

What's happening in the second example is that None do get evaluated at definition (it evaluates to None duh!), but the code that conditionally assigns [] to L only gets compiled and is run each time (the condition passes).

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