On Tue, 21 Jan 2014 20:11:02 +0100
Mû m...@melix.net 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
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
On Wed, Jan 22, 2014 at 6:11 AM, Mû m...@melix.net wrote:
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
Le 21/01/2014 20:19, Chris Angelico a écrit :
On Wed, Jan 22, 2014 at 6:11 AM, Mû m...@melix.net wrote:
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
On Wed, Jan 22, 2014 at 6:36 AM, Mû m...@melix.net wrote:
These were clear and quick answers to my problem. I did not think of this
possibility: the default argument is created once, but accessible only by
the function, therefore is not a global variable, whereas it looks like if
it were at
On Tuesday, January 21, 2014 9:46:16 PM UTC+2, Chris Angelico wrote:
On Wed, Jan 22, 2014 at 6:36 AM, Mû m...@melix.net wrote:
These were clear and quick answers to my problem. I did not think of this
possibility: the default argument is created once, but accessible only by
the function,
