Question

If I have a file called test1.lua

function print_hi()
   print("hi")
end

and I want to make the function available to another file called test2.lua, I write:

require 'test1'
function print_hi_and_bye()
   print_hi()
   print('bye')
end

But, now let’s say I have a third function called test3.lua to which I want to expose print_hi_and_bye() but NOT print_hi(). If I require ‘test2’ I will have access to both the print_hi and print_hi_and_bye() functions. How do I get around this and keep the functions of test1 local to test2 so that nothing else uses them by mistake? Is there a way to do this with lua’s loading facilities and not just by refactoring code?

Thanks

Answer 1

You need to make test1.lua functions only visible for whom requested it. For this, some changes in files test1.lua and test2.lua are needed:

test1.lua

local pkg = {}
function pkg.print_hi()
    print("hi")
end
return pkg

test2.lua

local m = require 'test1'
function print_hi_and_bye()
    m.print_hi()
    print('bye')
end

The changes are minimal, and now you can use the functions only in the files you request them.

In Lua 5.1, you can use the module function in test1.lua for convenience.

module("test1")

function print_hi()
    print("hi")
end

In Lua 5.2, this function is deprecated as it violated the design principles of Lua; instead you should do as shown in the first example.