why del is not a function or method?

[Thomas Jollans]

On 2017-10-17 03:07, Steve D'Aprano wrote:
> On Tue, 17 Oct 2017 08:57 am, Thomas Jollans wrote:
> 
>> On 16/10/17 21:12, Stefan Ram wrote:
>>> ram at zedat.fu-berlin.de (Stefan Ram) writes:
>>>> »x = None« observably has not the same effect as »del x«:
>>>
>>>   Paradoxically, /deleting/ a local variable which did not
>>>   ever exist, has the effect of creating a ghost of that
>>>   local variable which then will hide a global variable:
> 
> [...]
>>>   . Case 2: The »del x« creates a ghost »local variable 'x'«
>>>   which now will hide the global variable »x«:
>>>
>>> |>>> x = 9
>>> |>>> def f():
>>> |...     del x
>>> |...     print( x )
>>> |...
>>> |>>> f()
>>> |UnboundLocalError: local variable 'x' referenced before assignment
>>>
>>>   .
>>
>> That's not what happens. The UnboundLocalError is in the del statement,
>> not the print call:
> 
> That's exactly what happens! Stefan merely explains it poorly. There's no
> mysterious "ghost", it is just that x is treated as a local variable instead
> of a global, and you can't delete the local before it is bound to.
> 
> `del x` is treated as a form of assignment, just like `x = 123`, and makes `x`
> a local variable. So in this code:

You're right of course.

While in Stefan's example code the error *was* in the del statement, and
trying to delete something that's not either present in the local
namespace or explicitly "global" *will* fail, del *does* make a variable
local.


 >>> x = 1
 >>> def f():
 ...     try:
 ...         del x
 ...     except:
 ...         print('del exception caught')
 ...     print(x)
 ...
 >>> f()
 del exception caught
 Traceback (most recent call last):
   File "<stdin>", line 1, in <module>
   File "<stdin>", line 6, in f
 UnboundLocalError: local variable 'x' referenced before assignment
 >>> def g():
 ...     global x
 ...     del x
 ...
 >>> g()
 >>> x
 Traceback (most recent call last):
   File "<stdin>", line 1, in <module>
 NameError: name 'x' is not defined
 >>>


> 
> def f():
>     print(x)
> 
> there is no assignment to x, so it is treated as a global. But if we write:
> 
> def f():
>     x = 123
>     print(x)
> 
> the Python compiler sees the assignment to x and makes x a local variable.
> Note that the rules for this are deliberately very, very simple, so this will
> make x a local:
> 
> def f():
>     print(x)  # x is local
>     x = 123
> 
> and even this will too:
> 
> def f():
>     if False:
>         x = 123  # dead code
>     print(x)  # but enough to make x a local
> 
> 
> You can replace the line "x = 123" with "del x" in any of those functions, and
> you will get the same effect: x will be treated as a local.
> 
> 
>