paolo veronelli said:
I've a vague idea of the differences,I don't know scheme anyway.
I'd like to see an example to show what is missing in python about
closures
A long, long time ago, in a Python far away, there were three
namesapaces: local (to a function), global and builtin. When looking
up a name, Python would first search the local namespace (if there
were one, i.e. we're inside a function), if the name were not found,
it would search the global namespace, and, if still not found, fall
back on the builtin namespace.
(Let's forget about builtins, for this discussion. (You can always
explicitly access the builtins namespace via __builtins__.))
a = 3
def fn():
print a
fn() # => 3
No surprises there. "a" is found in the global namespace.
a = 3
def fn():
a = 4
print a # => 4
fn()
print a # => 3
Inside "fn" a local variable "a" is created (shadowing the global
one), it is printed, goes out of scope and disappears. Inside "fn" it
shadowed the global "a" ... which is why the global "a" is unaffected
by the "a = 4" inside "fn", and "3" was printed on the last line.
Now, let's make a tiny alteration to the last example:
a = 3
def fn():
print a # This is the added line
a = 4
print a
fn()
print a
Run this, and you'll get an error, on the line we added:
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "/usr/tmp/python-BxFkNE", line 7, in ?
fn()
File "/usr/tmp/python-BxFkNE", line 3, in fn
print a
UnboundLocalError: local variable 'a' referenced before assignment
Why?
Because the compiler decides that "a" refers to a local variable in
"fn" by noticing that there is an "a = ..." in the function body, and
therefore never looks for "a" outside the local scope. But the first
"print a" comes before that local variable is bound, so it's referring
to a variable which has not been bound yet.
This illustrates that the appearance of "a =" in a function body makes
python believe that "a" is local. But what if you want the function to
modify some global state? What if you want to modify the global "a" ?
For this reason, Python has the "global" keyword:
a = 3
def fn():
global a
print a # => 3
a = 4
print a # => 4
fn()
print a # => 4
All is perfect.
But then, one dark day, nested scopes were added to Python. What does
this mean? Well, in the good old times (say, python 1.5.2) you would
get the following behaviour:
a = 3
def outer():
a = 4
def inner():
print a
return inner
outer()() # => 3
Why? "a" is not found in the local scope of "inner", so Python looks
for it in the global scope, where it does find it ... and where it is
bound to "3".
But in a modern Python, the output of "outer()()" is "4", because
nowadays Python looks for the name in any enclosing function-local
scopes, before getting out to the global scope. So, in a modern
Python, the "a = 4" binding, which is local to "outer" is found.
Now, introducing a local binding of "a" in "inner", should offer no
surprises:
a = 3
def outer():
a = 4
def inner():
a = 5
print a
return inner
outer()() # =>
And adding the same "print a" we added to "fn" earlier ...
a = 3
def outer():
a = 4
def inner():
print a
a = 5
print a
return inner
outer()() # =>
.... gives us exactly the same error
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "/usr/tmp/python-BxFY2c", line 12, in ?
outer()()
File "/usr/tmp/python-BxFY2c", line 7, in inner
print a
UnboundLocalError: local variable 'a' referenced before assignment
We could try to fix it, as we did before, by adding a "global a"
declaration:
a = 3
def outer():
a = 4
def inner():
global a
print a # => 3
a = 5
print a # => 5
return inner
outer()() # =>
.... but this results in the intermediate local function scope (the one
of "outer") being ignored (as evidenced by the fact that the first
"print a" gives us "3" rather than "4"). So, it seems that nested
scopes were introduced into Python in a read-only fashion.
The problem is that there is no way of saying "mutate the binding for
the name, wherever you find it". You can say:
a) Mutate or create a binding in the builtin scope:
__builtins__.a = ...
b) Mutate or create a binding in the global scope:
global a
a = ...
c) Mutate or create a binding in the local scope:
a = ...
So, one solution might be to introduce a way of saying "Mutate the
binding, wherever you find it":
lexical a
a =
But, viewed from another perspective, the problem is that Python
offers no way to distinguish between mutating existing bindings and
creating new ones; both are spelt "=". If there were a way to specify
whether you mean "create a new binding" or "rebind the name", then
there would be no need for the "lexical" (or even the "global")
declaration.
This is what is done in some other languages. For example in Scheme:
- Find the nearest binding of "a" and rebind it to "3":
(set! a 3)
- Create a new (lexically nested) variable "a" and bind it to "3":
(let ((a 3) ...)
So, what is missing about Python closures, is the mutability of the
enclosed variables. There is a simple way of hacking around it: make
the enclosed variable a mutable, and mutate it, rather than trying to
rebind the name (thereby avoiding the "a =" syntax which tells Python
to create a local binding of "a") ...
def outer():
a = [4]
def inner():
print a[0]
a[0] = a[0] + 1
return inner
fn = outer()
fn() # => 4
fn() # => 5
fn() # => 6
It works, but it severely reduces the elegance and convenience of
closures. If you appreciate closures, you are likely to consider this
a big pain. If you don't appreciate closures you probabl can't see
what the fuss is about, and are likely to tell people that stateful
functions in Python should be implemented as methods of classes.
I happen to appreciate closures.
I don't know about closures in Ruby,