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Let's say I have file named test.m containing

test[arg1_] := ( Print[arg1];)

What is the best technique for calling test like test["value"] while preventing test from being added to the $ContextPath?

The following almost works except you must know function values ahead of time

Block[{$ContextPath,test}, Needs["test`"];test["test"]];

My question is: How do I generalize the call above to work with any number of functions without the user needing to input definitions manually?

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  • $\begingroup$ @Nasser although I agree generally speaking everything should be in package, let's pretend the example of above is about calling functions that exist in a notebook. Now surely you are not recommending me wrap the notebook functions with Package statements simply to call a notebooks function from another notebook. At the end of the day I believe such functionality could be useful and to me it really seems like a solution is to implausible. $\endgroup$
    – William
    Commented Aug 3, 2013 at 23:12

1 Answer 1

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As Nasser notes in a comment every Symbol has a context. You should be aware that Symbols are created during parsing. See: Local variables in Module leak into the Global context.

Alright, now that we worked out what you want here is the simplified answer:

We can perform a similar operation to BeginPackage with Block:

Block[{$ContextPath = {"runPrv`", "System`"}, $Context = "runPrv`"}, . . .]

We can combine this with Leonid's method from Is it possible to use Begin and End inside a Manipulate? to keep contexts from being fully resolved until we are ready for evaluation inside the Block. (Note that Global` Symbols are still created, as discussed above, but they will not be defined.) I believe "runPrv`" may be left out of $ContextPath in our application so long as we don't change the $Context from "runPrv`" within the Block itself. Finally we have:

SetAttributes[runPrivate, HoldAllComplete];

runPrivate[code_] :=
  With[{body = MakeBoxes @ code},
    Block[{$ContextPath = {"System`"}, $Context = "runPrv`"},
      ToExpression @ body]]

Now:

runPrivate[
  Get["test.m"];
  a = 5;
  test[a]
]

5

Global Symbols a and test remain undefined:

?a
?test

Global`a

Global`test

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  • $\begingroup$ +1 The 2nd example is great and helps clarify. This relates strongly to the original question. How would I modify this following code so a isn't defined. Block[{$ContextPath = {"test`", "System`"}, $Context = "test`"}, a = 1;] I see in your response here mathematica.stackexchange.com/questions/24347/… you only mention 2 approaches for avoiding to define declarations int the $Context. Is there a technique that avoids pre-pending the context test` and in addition it doesn't require you run Clear[function1] first. $\endgroup$
    – William
    Commented Aug 5, 2013 at 0:13
  • $\begingroup$ @Liam You are working against the normal evaluation order, and there may be other approaches depending on what you want, such as a Cell with a unique Context, but that specific case may best be handled as Leonid showed here, e.g.: Block[{$ContextPath = {"test`", "System`"}, $Context = "test`", code = MakeBoxes[a = 1]}, ToExpression[code]] $\endgroup$
    – Mr.Wizard
    Commented Aug 5, 2013 at 0:23
  • $\begingroup$ @Liam I put this example in the question. I think perhaps we finally have what you want? $\endgroup$
    – Mr.Wizard
    Commented Aug 5, 2013 at 0:36
  • $\begingroup$ Yes. Now this isn't exactly production quality, but I'm going to assume there isn't really a better way to suppress the message like z::shdw: Symbol z appears in multiple contexts {test`,Global`}; definitions in context test` may shadow or be shadowed by other definitions. >> except with Quiet[Block[{$ContextPath = {"test`", "System`"}, $Context = "test`", code = MakeBoxes[p = 1]}, ToExpression[code]], General::shdw] or is there another way? $\endgroup$
    – William
    Commented Aug 5, 2013 at 0:40
  • $\begingroup$ @Liam I was busy updating my answer. Could you please give me an example using the current runPrivate code that produces this shadowing when run in a fresh session? $\endgroup$
    – Mr.Wizard
    Commented Aug 5, 2013 at 0:49

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