Python中设置变量作为默认值时容易遇到的错误
思考一下下面的代码片段：
def foo(numbers=[]): numbers.append(9) print numbers
在这里，我们定义了一个 list （默认为空），给它加入9并且打印出来。
>>> foo() [9] >>> foo(numbers=[1,2]) [1, 2, 9] >>> foo(numbers=[1,2,3]) [1, 2, 3, 9]
看起来还行吧？可是当我们不输入number 参数来调用 foo 函数时，神奇的事情发生了：
>>> foo() # first time, like before [9] >>> foo() # second time [9, 9] >>> foo() # third time... [9, 9, 9] >>> foo() # WHAT IS THIS BLACK MAGIC?! [9, 9, 9, 9]
那么，这是神马情况？直觉告诉我们无论我们不输入 number 参数调用 foo 函数多少次，这里的9应该被分配进了一个空的 list。这是错的！在Python里，函数的默认值实在函数定义的时候实例化的，而不是在调用的时候。
那么我们仍然会问，为什么在调用函数的时候这个默认值却被赋予了不同的值？因为在你每次给函数指定一个默认值的时候，Python都会存储这个值。如果在调用函数的时候重写了默认值，那么这个存储的值就不会被使用。当你不重写默认值的时候，那么Python就会让默认值引用存储的值（这个例子里的numbers）。它并不是将存储的值拷贝来为这个变量赋值。这个概念可能对初学者来说，理解起来会比较吃力，所以可以这样来理解：有两个变量，一个是内部的，一个是当前运行时的变量。现实就是我们有两个变量来用相同的值进行交互，所以一旦 numbers 的值发生变化，也会改变Python里面保存的初始值的记录。
那么解决方案如下：
def foo(numbers=None):
if numbers is None:
numbers = []
numbers.append(9)
print numbers
通常，当人们听到这里，大家会问另一个关于默认值的问题。思考下面的程序：
def foo(count=0): count += 1 print count
当我们运行它的时候，其结果完全是我们期望的：
>>> foo() 1 >>> foo() 1 >>> foo(2) 3 >>> foo(3) 4 >>> foo() 1
这又是为啥呢？其秘密不在与默认值被赋值的时候，而是这个默认值本身。整型是一种不可变的变量。跟 list 类型不同，在函数执行的过程中，整型变量是不能被改变的。当我们执行 count+=1 这句话时，我们并没有改变 count 这个变量原有的值。而是让 count 指向了不同的值。可是，当我们执行 numbers.append(9) 的时候，我们改变了原有的 list 。因而导致了这种结果。
下面是在函数里使用默认值时会碰到的另一种相同问题：
def print_now(now=time.time()): print now
跟前面一样，time.time() 的值是可变的，那么它只会在函数定义的时候计算，所以无论调用多少次，都会返回相同的时间 — 这里输出的时间是程序被Python解释运行的时间。
>>> print_now() 1373121487.91 >>> print_now() 1373121487.91 >>> print_now() 1373121487.91
* 这个问题和它的解决方案在 Python 2.x 和 3.x 里都是类似的，在Python 3.x 里面唯一的不同，是里面的print 表达式应该是函数调用的方式（print(numbers)）。
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