“main” function in Lua?

Question

In python, one would usually define a main function, in order to allow the script to be used as module (if needed):

def main():
    print(\"Hello world\")
           


        

Answer 1

I am going to suggest yet another variation, that seems to work on lua5.1 as well as lua5.2:

function is_main(offset)
    return debug.getinfo(4 + (offset or 0)) == nil
end

if is_main() then
    print("Main chunk!")
else
    print("Library chunk!")
end

If you do not feel like defining an extra function is_main for this purpose, you can just do:

if debug.getinfo(3) == nil then
    print("Main chunk!")
else
    print("Library chunk!")
end

Answer 2

if arg ~= nil and arg[0] == string.sub(debug.getinfo(1,'S').source,2) then
  print "Main file"
else
  print "Library"
end

An explanation:

1. Lua calls a script from the interpreter with the arg table, with arg[0] being the name of the script.
2. The debug.getinfo function returns a table of information describing the function at the level of the stack given by its numbered argument (1 refers to the function calling getinfo; 0 refers to getinfo itself). Its second parameter is optional: it specifies which subset of the getinfo fields to retrieve. The 'S' defines Source names.
3. The source field of the table returned by getinfo for files contains the name of the file the function was defined in, prefixed with an @. By passing the value of that field to string.sub(...,2), we strip that @. (The short_src field contains the name without the @, but that's because it's defined as a "print-friendly" name, and is less safe than stripping source.)

Note that this code uses a debug function, so in 5.2 you'll have to require debug to be able to use it.