[Tutor] Why is it invalid syntax to have a particular dictionary value as an argument?

[Dave Angel]

On 04/07/2015 10:16 PM, boB Stepp wrote:
>
> Despite Mark's warning, I feel I must see if I understand what is going on here.
>
> Switching to Py 3.4 since I am now at home:
>
> Python 3.4.3 (v3.4.3:9b73f1c3e601, Feb 24 2015, 22:44:40) [MSC v.1600
> 64 bit (AMD64)] on win32
> Type "copyright", "credits" or "license()" for more information.
>>>> d = {'a': '123'}
>>>> def func(s=d['a']):
>                print(s)
>                print(d['a'])
>
>>>> func()
> 123
> 123
>>>> d['a'] = 'new value'
>>>> func()
> 123
> new value
>
> I added an additional print to the function to show the dictionary
> entry's behavior.
>
> First, my current understanding is that this form of the function does
> not object to the presence of d['a'] in its parameter list because s
> is the real parameter, d['a'] is its default value, but s is not
> actually evaluated until run time.

s is not evaluated till the print statement.  s is *bound* at function 
call time.  And at that time it is either bound to the object passed by 
the caller, or to the default object.

In a simple assignment statement:
    a = b + 6

the expression on the right is evaluated.  The name on the left is not 
evaluated, it is bound to.  So we say
    "a is bound to the result of the expression b+6"

>
> But once s *is* evaluated, it stores  a reference to the original

   s is bound to the expression ''default_object'', which is to say it 
copies the same reference that the default object stored earlier.  So it 
is bound to '123'

> object, '123'. Changing d['a'] outside the function to a new value
> does not alter the fact that s is storing the very same reference to
> '123'. After reassigning d['a'] to point to the new object 'new
> value', a new call to func() shows s still referencing the original
> object and d['a'] referencing the new object. Is my comprehension of
> these details correct? If yes, this is why I must constantly remind
> myself that identifiers store references to objects, and that some
> objects are mutable and some aren't, and these Python facts of life
> are constantly challenging my old FORTRAN <= 77 ways of thinking...
> ~(:>))
>


-- 
DaveA