Will the parameters be initialized when Python uses list as a function parameter?

See this section of code:

def foo (A, b=[]):    b.append (a)    print B

Reply content:

>>>  def  foo   ( bar  =  []):       ...  return  bar  >>>  foo   func_name   ' foo '  >>>  foo   func_defaults   ([],)  >>>  foo   ()  is  foo   func_defaults  [ 0  ]  true   

Official documents are explained here: Default args is evaluated at the time of definition, only once
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But......

>>> def f(a, b=[]):...     b.append(a)...     print b... >>> f(1)[1]>>> f(1)[1, 1]>>> def f(a, b=None):...     b = b if b is not None else []...     b.append(a)...     

Don't say anything, add an ID () output B's so-called address, you understand

No, def foo (a=[]) This function parameter is called the default value of the parameter, only once the function declaration is initialized. And then it won't change any more.

Note, it is recommended to know the difference between Def foo (a=[]) and foo (a=[]): The former is the default value of the parameter, and the latter is the keyword arguments. There are subtle differences between this def foo (*args, **kargs) and this foo (*args, **kargs). No, the default values are shared, created only once, and do not create a new object every time. This means that using a Mutable object as the default value of a function can cause confusion in the function. Similarly, using a dictionary as the default parameter results in a similar return.

deffoo(k,v,fdict={}):    fdict[k]=v    printfdictfoo(1,2)foo(3,4)

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