Why can't I pass a direct reference to a dictionary value to a function?

Question

Earlier today I asked a question about passing dictionary values to a function. While I understand now how to accomplish what I was trying to accomplish the why question (which was not asked) was never answered. So my follow up is why can't I

def myFunction(newDict['bubba']):
    some code to process the parameter

Is it simply because the parser rules do not allow this? I Googled for +Python +function +"allowable parameters" and did not find anything useful so I would appreciate any information.

I am oversimplifying what is going on. I have a dictionary that is structured like

myDict={outerkey1:(innerkey1:value,innerkey2:value,. . .innerkeyn:value),outerkey2:(innerkey1:value,innerkey2:value,. . .innerkeyn:value),. . .}

As I said, I know how to do what I wanted-I got a very helpful answer. But I started wondering why

 def myFunction(outerkey,myDict[outerkey]):

gives a syntax error which finely it occurred to me that it has to be a parsing problem.

Answer 1

Yes, the parser will reject this code.

Parameter lists are used in function definitions to bind identifiers within the function to arguments that are passed in from the outside on invocation.

Since newDict['bubba'] is not a valid identifier, this doesn't make any sense -- you need to provide it as an invocation argument instead of a function parameter, since function parameters can only be identifiers.

Because you seem interested in the formal grammar, here are the relevant parts:

funcdef ::= 
             [decorators] "def" funcname "(" [parameter_list] ")"
              ":" suite

parameter_list ::= 
                 (defparameter ",")*
                (~~"*" identifier [, "**" identifier]
                 | "**" identifier
                 | defparameter [","] )

defparameter ::= 
             parameter ["=" expression]

identifier ::= 
             (letter|"_") (letter | digit | "_")*

In fact, the construct you are trying to use as an identifier is a subscription:

subscription ::= 
             primary "[" expression_list "]"

Answer 2

It looks like you might be confused between the definition of a function and calling that function. As a simple example:

def f(x):
    y = x * x
    print "x squared is", y

By itself, this function doesn't do anything (ie. it doesn't print anything). If you put this in a source file and run it, nothing will be output. But then if you add:

f(5)

you will get the output "x squared is 25". When Python encounters the name of the function followed by actual parameters, it substitutes the value(s) of those actual parameters for the formal parameter(s) in the function definition. In this case, the formal parameter is called x. Then Python runs the function f with the value of x as 5. Now, the big benefit is you can use the same function again with a different value:

f(10)

prints "x squared is 100".

I hope I've understood the source of your difficulty correctly.

Answer 3

You can only pass references to objects.

a = { 'a' : 123, 'b' : [ 1, 2, 3 ] }

def f(dictionary_value):
    dictionary_value.append(4)

f( a['b'] )

Does what you expect. Dictionary element a['b'] is passed "by reference" to the function f which updates the dictionary element.