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Re: How to detect that a function argument is the default one

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----- Original Message -----
> From: "ast" <nomail@invalid.com>
> I have the idea to write:
> 
> def __init__(center=(0,0), radius=10, mass=None)):
> 
>     if  mass == None:
>         self.mass = radius**2
>     else:
>         self.mass = mass
> 
> but maybe Python provides something clever.
> 
> Thx

If you like one-liners,

def __init__(self, center=(0,0), radius=10, mass=None):
    self.center = center
    self.radius = radius
    self.mass = (mass is None and radius**2) or mass

But there's an issue with that solution: self.mass being computed during the instance initialization, what if radius is changing ?

c1 = Circle((0,0), 10, None)
print c1.mass
20
c1.radius = 20
print c1.mass
20 # that is unexpected

Everytime an attribute is computed from another attribute, it should ring a python bell : property

def __init__(self, center, radius, mass):
    self.center = center
    self.radius = radius
    self._mass = mass

    @property
    def mass(self):
         return self._mass if self._mass is not None else self.radius*2


c1 = Circle((0,0), 10, None)
print c1.mass
20
c1.radius = 20
print c1.mass
40

JM

Note : what is the mass of a circle ?


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Re: How to detect that a function argument is the default one Jean-Michel Pichavant <jeanmichel@sequans.com> - 2014-12-10 17:10 +0100
  Re: How to detect that a function argument is the default one Jussi Piitulainen <jpiitula@ling.helsinki.fi> - 2014-12-10 19:20 +0200

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