Behavior of Remove inside a Package

Question

I have a notebook containing (in a Code cell, Initialization Group)

BeginPackage["aPackage`"];
aTestFunction::usage = "aTestFunction usage";
Begin["`Private`"];
(* Remove[aTestFunction]; *)
aTestFunction[arg_] := arg + 1;
Protect[aTestFunction];
End[];
EndPackage;

and further

Needs["aPackage`", NotebookFileName[]];
?aTestFunction
aTestFunction[1]
Quit[];

All that honestly earns its pay. But if I de - comment the Remove command, it produces.

Information::notfound: "Symbol aTestFunction not found."

and I can' t realize the reason. All things considered, Remove removes the aTestFunction symbol (provided that it exists) from the aPackage context but should not prevent it from being defined at a later time. It sounds quite different from what one can observe in the Global context: this

Context[]
a = 1;
Remove[a];
a = 2;
a + 1

returns

Global
3

.

closely related: http://mathematica.stackexchange.com/q/114769/5478 — Kuba, May 03 '17 at 20:45

score 3 · Accepted Answer · edited Apr 13 '17 at 12:55

Shortly:

Here what is going on, aTestFunction was created in aPackage` context, then you've removed it and since the current context is aPackage`Private` the definition of your function is created there.

More precisely:

New symbols are created in current $Context during read time, (1) the existance of aTestFunction between BeginPackage and Begin lines makes the aTestFunction an exported symbol (that means in aPackage`, because the symbol was not found and the current $Context is apackage`).

After (2) Remove it's no longer there and (3) your definition is read in aPackage`Private` context. But aPackage`Private` is not going to be on $ContextPath or $Context after EndPackage[] so it is not found by ?.

I'm strongly encouraging to read a very nice answer of Szabolcs:

How symbol lookup actually works

BeginPackage["aPackage`"]; 
  aTestFunction::usage = "aTestFunction usage"; (*1*)
Begin["`Private`"];
  Remove[aTestFunction]; (*2*)
  aTestFunction[arg_] := arg + 1; (*3*)
  Protect[aTestFunction];
End[];
EndPackage;

? aTestFunction

Information::notfound: Symbol aTestFunction not found. >>

? aPackage`Private`aTestFunction

Hello ! Thanks for your reply ! It's all clear now. I think that my oversight was in leaving behind that usage command is totally different from a simple string assignment. — mitochondrial, Apr 08 '16 at 13:59

Behavior of Remove inside a Package

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