Certainly the way default arguments work with mutable types is not the most
intuitive and I think your complaint has some merit.
However how would you define the following to work:
def foo():
cons = [set(), [], (),]
funs = []
for ds in cons:
def g(arg:=ds):
return arg
funs.append(g)
return funs
How would you evaluate "ds" in the context of the call?
If it were to have the same observable behavior as def g(arg=ds) except
that you would get "fresh" reference on each invocation you would get the
following:
assert [f() for f in foo()] == [set(), [], ()]
Note it cannot be a simple syntactic transform because:
class _MISSING: pass
def foo():
cons = [set(), [], (),]
funs = []
for ds in cons:
def g(arg=_MISSING):
if arg is _MISSING:
arg = eval('ds') # equivalent to arg = ds so does not
produce a fresh reference
return arg
funs.append(g)
return funs
assert [f() for f in foo()] == [(), (), ()]
Granted the way closures work (especially in the context of loops) is also
a pretty unintuitive, but stands as a barrier to easily implementing your
desired behavior.
And even if that wasn't the case we still have the issue that eval('ds')
doesn't give you a fresh reference.
Would it implicitly deepcopy ds? e.g.:
class _MISSING: pass
def build_g(default):
def g(arg=_MISSING):
if arg is _MISSING:
arg = deepcopy(default)
return arg
return g
def foo():
cons = [set(), [], (),]
funs = []
for ds in cons:
g = build_g(ds)
funs.append(g)
return funs
What if ds doesn't implement __deepcopy__?
On Mon, May 18, 2020 at 7:11 AM Richard Damon <rich...@damon-family.org>
wrote:
> On 5/18/20 9:06 AM, James Lu wrote:
> > "There should be one-- and preferably only one --obvious way to do it."
>
> *obvious*
>
> multiple ways are allowed as long as there is one clear preference.
>
> --
> Richard Damon
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