[Python-ideas] Continuation of `__name__` or a builtin function for general name getting

[MRAB]

On 2017-06-18 22:38, Alireza Rafiei wrote:
> Hi all,
> 
> I'm not sure whether this idea has been discussed before or not, so I 
> apologize in advanced if that's the case.
> 
> Consider the behavior:
> 
>      >>> f = lambda: True
>      >>> f.__name__
>     '<lambda>'
>      >>> x = f
>      >>> x.__name__
>     '<lambda>'
> 
> 
> I'm arguing the behavior above is too brittle/limited and, considering 
> that the name of the attribute is `__name__`, not entirely consistent 
> with Python's AST. Consider:
> 
>      >>> f = lambda: True
>      >>> x = f
> 
> 
> At the first line, an ast.Assign would be created whose target is an 
> ast.Name whose `id` is `f`.
> At the second line, an ast.Assign would be created whose target is an 
> ast.Name whose `id` is `x`.
> However, as you can see `__name__` special method returns 'lambda' in 
> both cases (just like it was defined 
> https://docs.python.org/3/library/stdtypes.html#definition.__name__), 
> whereas I think either it should have returned '<lambda>' and 'x' or a 
> new function/attribute should exist that does so and more.
> 
> For example, consider:
> 
>      >>> x_1 = 1
>      >>> x_2 = 1
>      >>> x_3 = 1
>      >>> x_4 = x_1
>      >>> for i in [x_1, x_2, x_3, x_4]:
>      >>>     print(i)
>     1
>     1
>     1
>     1
> 
> 
> Now assume such a function exist and is called `name`. Then:
> 
>      >>> name(1)
>     '1'
>      >>> name("Something")
>     "Something"
>      >>> name(x_1)
>     'x_1'
>      >>> name(x_4)
>     'x_4'
>      >>> name(x_5)
>     'x_5' # Or an Exception!
>      >>> def itername(collection):
>      >>>     for i in map(lambda x: name(x), collection):
>      >>>         yield i
>      >>>
>      >>> for i in [x_1, x_2, x_3, x_4]:
>      >>>     print(i, name(i))
>     1, 'i'
>     1, 'i'
>     1, 'i'
>     1, 'i'
>      >>> for i in itername([x_1, x_2, x_3, x_4]):
>      >>>     print(i)
>     'x_1'
>     'x_2'
>     'x_3'
>     'x_4'
> 
[snip]

That's not correct.

Look at the definition of 'itername'. The lambda returns the result of 
name(x), which is 'x'.

Therefore, the correct result is:

'x'
'x'
'x'
'x'