tl;dr version:
if !isdefined(:__init__) || Base.function_module(__init__) != MyModule
main()
end
Explanation:
There seems to be some confusion. Python and Julia work very differently in terms of their "modules" (even though the two use the same term, in principle they are different).
In python, a source file is either a module or a script, depending on how you chose to "load" / "run" it: the boilerplate exists to detect the environment in which the source code was run, by querying the __name__ of the embedding module at the time of execution. E.g. if you have a file called mymodule.py, it you import it normally, then within the module definition the variable __name__ automatically gets set to the value mymodule; but if you ran it as a standalone script (effectively "dumping" the code into the "main" module), the __name__ variable is that of the global scope, namely __main__. This difference gives you the ability to detect how a python file was ran, so you could act slightly differently in each case, and this is exactly what the boilerplate does.
In julia, however, a module is defined explicitly as code. Running a file that contains a module declaration will load that module regardless of whether you did using or include; however in the former case, the module will not be reloaded if it's already on the workspace, whereas in the latter case it's as if you "redefined" it.
Modules can have initialisation code via the special __init__() function, whose job is to only run the first time a module is loaded (e.g. when imported via a using statement). So one thing you could do is have a standalone script, which you could either include directly to run as a standalone script, or include it within the scope of a module definition, and have it detect the presence of module-specific variables such that it behaves differently in each case. But it would still have to be a standalone file, separate from the main module definition.
If you want the module to do stuff, that the standalone script shouldn't, this is easy: you just have something like this:
module MyModule
__init__() = # do module specific initialisation stuff here
include("MyModule_Implementation.jl")
end
If you want the reverse situation, you need a way to detect whether you're running inside the module or not. You could do this, e.g. by detecting the presence of a suitable __init__() function, belonging to that particular module. For example:
### in file "MyModule.jl"
module MyModule
export fun1, fun2;
__init__() = print("Initialising module ...");
include("MyModuleImplementation.jl");
end
### in file "MyModuleImplementation.jl"
fun1(a,b) = a + b;
fun2(a,b) = a * b;
main() = print("Demo of fun1 and fun2. \n" *
" fun1(1,2) = $(fun1(1,2)) \n" *
" fun2(1,2) = $(fun2(1,2)) \n");
if !isdefined(:__init__) || Base.function_module(__init__) != MyModule
main()
end
If MyModule is loaded as a module, the main function in MyModuleImplementation.jl will not run.
If you run MyModuleImplementation.jl as a standalone script, the main function will run.
So this is a way to achieve something close to the effect you want; but it's very different to saying running a module-defining file as either a module or a standalone script; I don't think you can simply "strip" the module instruction from the code and run the module's "contents" in such a manner in julia.
