Function defs with mutable arguments hold a reference to the mutable
container such that all invocations access the same changeable container.
To get separate mutable default arguments, use:
def f(x=None):
if x is None: x=[2,3]
Emile
On 01/21/2014 11:11 AM, Mû wrote:
Hi everybody,
A friend of mine asked me a question about the following code:
[code]
def f(x=[2,3]):
x.append(1)
return x
print(f())
print(f())
print(f())
[/code]
The results are [2, 3, 1], [2, 3, 1, 1] and [2, 3, 1, 1, 1].
The function acts as if there were a global variable x, but the call of
x results in an error (undefined variable). I don't understand why the
successive calls of f() don't return the same value: indeed, I thought
that [2,3] was the default argument of the function f, thus I expected
the three calls of f() to be exactly equivalent.
I'm don't know much about python, does anybody have a simple explanation
please?
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