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，我遇到了这样的事情.

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=[] 只被调用一次，fun1() 的输出是美好的.但是为什么 L=None 每次都在 fun2() 中被调用.

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

当你定义一个函数时，默认参数的值会被评估，但函数体只会被编译.您可以通过属性检查函数定义的结果.有一个包含默认值的 __defaults__ 属性和包含主体的 __code__ 属性(因此这些是在定义函数时创建的).

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).

在第二个例子中发生的事情是 None 确实在定义时被评估(它评估为 None 废话！)，但是有条件地分配 [] 到 L 只被编译并每次运行(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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